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Abstract Linear Algebra ISingular Value Decomposition (SVD)

Linear Algebra. Session 9

Dr. Marco A Roque Sol

10 / 25 / 2018

Dr. Marco A Roque Sol Linear Algebra. Session 9

Complex EigenvaluesRepeated EigenvaluesDiagonalization

Complex Eigenvalues

In this section we consider again a system of n linear homogeneousequations with constant coefficients

X′ = AX

where the coefficient matrix A is real-valued. If we seek solutionsof the form x = veλt , then it follows that λ must be an eigenvalueand v a corresponding eigenvector of the coefficient matrix A.

In the case, λ is complex, we have complex eigenvalues andeigenvectors always appear in complex-conjugate. Thus, if wehave that

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

In this section

we consider again a system of n linear homogeneousequations with constant coefficients

X′ = AX

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

In this section we consider again

a system of n linear homogeneousequations with constant coefficients

X′ = AX

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

In this section we consider again a system of n

linear homogeneousequations with constant coefficients

X′ = AX

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

In this section we consider again a system of n linear homogeneousequations

with constant coefficients

X′ = AX

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

the coefficient matrix A is real-valued. If we seek solutionsof the form x = veλt , then it follows that λ must be an eigenvalueand v a corresponding eigenvector of the coefficient matrix A.

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

where the coefficient matrix

A is real-valued. If we seek solutionsof the form x = veλt , then it follows that λ must be an eigenvalueand v a corresponding eigenvector of the coefficient matrix A.

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

where the coefficient matrix A

is real-valued. If we seek solutionsof the form x = veλt , then it follows that λ must be an eigenvalueand v a corresponding eigenvector of the coefficient matrix A.

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

where the coefficient matrix A is real-valued.

If we seek solutionsof the form x = veλt , then it follows that λ must be an eigenvalueand v a corresponding eigenvector of the coefficient matrix A.

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

where the coefficient matrix A is real-valued. If we seek

solutionsof the form x = veλt , then it follows that λ must be an eigenvalueand v a corresponding eigenvector of the coefficient matrix A.

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

where the coefficient matrix A is real-valued. If we seek solutions

of the form x = veλt , then it follows that λ must be an eigenvalueand v a corresponding eigenvector of the coefficient matrix A.

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

where the coefficient matrix A is real-valued. If we seek solutionsof the form

x = veλt , then it follows that λ must be an eigenvalueand v a corresponding eigenvector of the coefficient matrix A.

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

where the coefficient matrix A is real-valued. If we seek solutionsof the form x = veλt ,

then it follows that λ must be an eigenvalueand v a corresponding eigenvector of the coefficient matrix A.

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

where the coefficient matrix A is real-valued. If we seek solutionsof the form x = veλt , then

it follows that λ must be an eigenvalueand v a corresponding eigenvector of the coefficient matrix A.

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

where the coefficient matrix A is real-valued. If we seek solutionsof the form x = veλt , then it follows that

λ must be an eigenvalueand v a corresponding eigenvector of the coefficient matrix A.

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

where the coefficient matrix A is real-valued. If we seek solutionsof the form x = veλt , then it follows that λ must be

an eigenvalueand v a corresponding eigenvector of the coefficient matrix A.

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

where the coefficient matrix A is real-valued. If we seek solutionsof the form x = veλt , then it follows that λ must be an eigenvalueand

v a corresponding eigenvector of the coefficient matrix A.

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

where the coefficient matrix A is real-valued. If we seek solutionsof the form x = veλt , then it follows that λ must be an eigenvalueand v

a corresponding eigenvector of the coefficient matrix A.

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

where the coefficient matrix A is real-valued. If we seek solutionsof the form x = veλt , then it follows that λ must be an eigenvalueand v a corresponding eigenvector

of the coefficient matrix A.

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

where the coefficient matrix A is real-valued. If we seek solutionsof the form x = veλt , then it follows that λ must be an eigenvalueand v a corresponding eigenvector of the

coefficient matrix A.

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

In the case,

λ is complex, we have complex eigenvalues andeigenvectors always appear in complex-conjugate. Thus, if wehave that

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

In the case, λ

is complex, we have complex eigenvalues andeigenvectors always appear in complex-conjugate. Thus, if wehave that

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

In the case, λ is complex,

we have complex eigenvalues andeigenvectors always appear in complex-conjugate. Thus, if wehave that

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

In the case, λ is complex, we have

complex eigenvalues andeigenvectors always appear in complex-conjugate. Thus, if wehave that

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

In the case, λ is complex, we have complex eigenvalues and

eigenvectors always appear in complex-conjugate. Thus, if wehave that

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

In the case, λ is complex, we have complex eigenvalues andeigenvectors

always appear in complex-conjugate. Thus, if wehave that

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

In the case, λ is complex, we have complex eigenvalues andeigenvectors always appear

in complex-conjugate. Thus, if wehave that

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

In the case, λ is complex, we have complex eigenvalues andeigenvectors always appear in complex-conjugate.

Thus, if wehave that

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

In the case, λ is complex, we have complex eigenvalues andeigenvectors always appear in complex-conjugate. Thus,

if wehave that

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

λ±k = µ± i ν;

v±k = a± i b

Complex Eigenvalues

X′ = AX

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

are two complex-conjugate eigenvalues and eigenvectors of thematrix A, then

X±(t) = e(µ±i ν)t (a± i b)

are complex-valued solutions, but taking in account that

e(µ±i ν)t = eµt (cos(νt)± i sin(νt))

and the principle of superposition, then we have that

X1(t) =1

(X+(t) + X−(t)

)X2(t) =

(X+(t)− X−(t)

Complex Eigenvalues

are two complex-conjugate

eigenvalues and eigenvectors of thematrix A, then

X±(t) = e(µ±i ν)t (a± i b)

X1(t) =1

(X+(t) + X−(t)

)X2(t) =

(X+(t)− X−(t)

Complex Eigenvalues

are two complex-conjugate eigenvalues and eigenvectors

of thematrix A, then

X±(t) = e(µ±i ν)t (a± i b)

X1(t) =1

(X+(t) + X−(t)

)X2(t) =

(X+(t)− X−(t)

Complex Eigenvalues

are two complex-conjugate eigenvalues and eigenvectors of thematrix A,

X±(t) = e(µ±i ν)t (a± i b)

X1(t) =1

(X+(t) + X−(t)

)X2(t) =

(X+(t)− X−(t)

Complex Eigenvalues

X±(t) = e(µ±i ν)t (a± i b)

X1(t) =1

(X+(t) + X−(t)

)X2(t) =

(X+(t)− X−(t)

Complex Eigenvalues

X±(t) = e(µ±i ν)t (a± i b)

X1(t) =1

(X+(t) + X−(t)

)X2(t) =

(X+(t)− X−(t)

Complex Eigenvalues

X±(t) = e(µ±i ν)t (a± i b)

are complex-valued

solutions, but taking in account that

X1(t) =1

(X+(t) + X−(t)

)X2(t) =

(X+(t)− X−(t)

Complex Eigenvalues

X±(t) = e(µ±i ν)t (a± i b)

are complex-valued solutions,

but taking in account that

X1(t) =1

(X+(t) + X−(t)

)X2(t) =

(X+(t)− X−(t)

Complex Eigenvalues

X±(t) = e(µ±i ν)t (a± i b)

are complex-valued solutions, but

taking in account that

X1(t) =1

(X+(t) + X−(t)

)X2(t) =

(X+(t)− X−(t)

Complex Eigenvalues

X±(t) = e(µ±i ν)t (a± i b)

X1(t) =1

(X+(t) + X−(t)

)X2(t) =

(X+(t)− X−(t)

Complex Eigenvalues

X±(t) = e(µ±i ν)t (a± i b)

e(µ±i ν)t =

eµt (cos(νt)± i sin(νt))

X1(t) =1

(X+(t) + X−(t)

)X2(t) =

(X+(t)− X−(t)

Complex Eigenvalues

X±(t) = e(µ±i ν)t (a± i b)

X1(t) =1

(X+(t) + X−(t)

)X2(t) =

(X+(t)− X−(t)

Complex Eigenvalues

X±(t) = e(µ±i ν)t (a± i b)

and the principle of superposition,

then we have that

X1(t) =1

(X+(t) + X−(t)

)X2(t) =

(X+(t)− X−(t)

Complex Eigenvalues

X±(t) = e(µ±i ν)t (a± i b)

X1(t) =1

(X+(t) + X−(t)

)X2(t) =

(X+(t)− X−(t)

Complex Eigenvalues

X±(t) = e(µ±i ν)t (a± i b)

X1(t) =1

(X+(t) + X−(t)

X2(t) =1

(X+(t)− X−(t)

Complex Eigenvalues

X±(t) = e(µ±i ν)t (a± i b)

X1(t) =1

(X+(t) + X−(t)

)X2(t) =

(X+(t)− X−(t)

Complex Eigenvalues

are two (real) solutions !!!

X1(t) = eµt (acos(νt)− bsin(νt))

X2(t) = eµt (acos(νt) + bsin(νt))

Complex Eigenvalues

are two

(real) solutions !!!

Complex Eigenvalues

Example 9.6

Solve the following ODE

x′ = Ax =

3 1 10 2 10 −1 2

Solution

Let’s find the eigenvalues of the matrix A

Complex Eigenvalues

Example 9.6

x′ = Ax =

3 1 10 2 10 −1 2

Solution

Complex Eigenvalues

Example 9.6

the following ODE

x′ = Ax =

3 1 10 2 10 −1 2

Solution

Complex Eigenvalues

Example 9.6

x′ = Ax =

3 1 10 2 10 −1 2

Solution

Complex Eigenvalues

Example 9.6

x′ = Ax =

3 1 10 2 10 −1 2

Solution

Complex Eigenvalues

Example 9.6

x′ = Ax =

3 1 10 2 10 −1 2

Solution

Complex Eigenvalues

Example 9.6

x′ = Ax =

3 1 10 2 10 −1 2

Solution

Complex Eigenvalues

Example 9.6

x′ = Ax =

3 1 10 2 10 −1 2

Solution

Let’s find

the eigenvalues of the matrix A

Complex Eigenvalues

Example 9.6

x′ = Ax =

3 1 10 2 10 −1 2

Solution

Let’s find the eigenvalues

of the matrix A

Complex Eigenvalues

Example 9.6

x′ = Ax =

3 1 10 2 10 −1 2

Solution

Let’s find the eigenvalues of the matrix

Complex Eigenvalues

Example 9.6

x′ = Ax =

3 1 10 2 10 −1 2

Solution

Complex Eigenvalues

|A− λI| =

∣∣∣∣∣∣3− λ 1 1

0 2− λ 10 −1 2− λ

∣∣∣∣∣∣ = 0

(3− λ)

∣∣∣∣2− λ 1−1 2− λ

∣∣∣∣ =

(3− λ)(λ2 − 4λ+ 5) = 0 =⇒

Complex Eigenvalues

|A− λI| =

∣∣∣∣∣∣3− λ 1 1

0 2− λ 10 −1 2− λ

∣∣∣∣∣∣ = 0

(3− λ)

∣∣∣∣2− λ 1−1 2− λ

∣∣∣∣ =

(3− λ)(λ2 − 4λ+ 5) = 0 =⇒

Complex Eigenvalues

|A− λI| =

∣∣∣∣∣∣3− λ 1 1

0 2− λ 10 −1 2− λ

∣∣∣∣∣∣ = 0

(3− λ)

∣∣∣∣2− λ 1−1 2− λ

∣∣∣∣ =

(3− λ)(λ2 − 4λ+ 5) = 0 =⇒

Complex Eigenvalues

|A− λI| =

∣∣∣∣∣∣3− λ 1 1

0 2− λ 10 −1 2− λ

∣∣∣∣∣∣ = 0

(3− λ)

∣∣∣∣2− λ 1−1 2− λ

∣∣∣∣ =

(3− λ)(λ2 − 4λ+ 5) = 0 =⇒

Complex Eigenvalues

|A− λI| =

∣∣∣∣∣∣3− λ 1 1

0 2− λ 10 −1 2− λ

∣∣∣∣∣∣ = 0

(3− λ)

∣∣∣∣2− λ 1−1 2− λ

∣∣∣∣ =

(3− λ)(λ2 − 4λ+ 5) = 0 =⇒

Complex Eigenvalues

λ1 = 2, λ2,3 =4±

√16− (4)(5)

2= 2± i

If λ1 = 3, then

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

0 1 10 −1 10 −1 −1

v1v2v3

0 1 10 0 20 0 0

v1v2v3

Complex Eigenvalues

λ1 = 2,

λ2,3 =4±

√16− (4)(5)

2= 2± i

If λ1 = 3, then

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

0 1 10 −1 10 −1 −1

v1v2v3

0 1 10 0 20 0 0

v1v2v3

Complex Eigenvalues

λ1 = 2, λ2,3 =4±

√16− (4)(5)

If λ1 = 3, then

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

0 1 10 −1 10 −1 −1

v1v2v3

0 1 10 0 20 0 0

v1v2v3

Complex Eigenvalues

λ1 = 2, λ2,3 =4±

√16− (4)(5)

2= 2± i

If λ1 = 3, then

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

0 1 10 −1 10 −1 −1

v1v2v3

0 1 10 0 20 0 0

v1v2v3

Complex Eigenvalues

λ1 = 2, λ2,3 =4±

√16− (4)(5)

2= 2± i

If λ1 = 3, then

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

0 1 10 −1 10 −1 −1

v1v2v3

0 1 10 0 20 0 0

v1v2v3

Complex Eigenvalues

λ1 = 2, λ2,3 =4±

√16− (4)(5)

2= 2± i

If λ1 = 3, then

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

0 1 10 −1 10 −1 −1

v1v2v3

0 1 10 0 20 0 0

v1v2v3

Complex Eigenvalues

λ1 = 2, λ2,3 =4±

√16− (4)(5)

2= 2± i

If λ1 = 3, then

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

0 1 10 −1 10 −1 −1

v1v2v3

0 1 10 0 20 0 0

v1v2v3

Complex Eigenvalues

λ1 = 2, λ2,3 =4±

√16− (4)(5)

2= 2± i

If λ1 = 3, then

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

0 1 10 −1 10 −1 −1

v1v2v3

0 1 10 0 20 0 0

v1v2v3

Complex Eigenvalues

λ1 = 2, λ2,3 =4±

√16− (4)(5)

2= 2± i

If λ1 = 3, then

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

0 1 10 −1 10 −1 −1

v1v2v3

0 1 10 0 20 0 0

v1v2v3

Complex Eigenvalues

and a corresponding eigenvector is

v(1) =

If λ2 = 2 + i , then

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

Complex Eigenvalues

a corresponding eigenvector is

v(1) =

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

Complex Eigenvalues

and a corresponding

eigenvector is

v(1) =

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

Complex Eigenvalues

v(1) =

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

Complex Eigenvalues

v(1) =

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

Complex Eigenvalues

v(1) =

λ2 = 2 + i , then

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

Complex Eigenvalues

v(1) =

If λ2 = 2 + i ,

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

Complex Eigenvalues

v(1) =

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

Complex Eigenvalues

v(1) =

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

Complex Eigenvalues

v(1) =

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

Complex Eigenvalues

3− (2 + i) 1 10 2− (2 + i) 10 −1 2− (2 + i)

v1v2v3

1− i 1 10 −i 10 −1 −i

v1v2v3

1− i 1 10 −i 10 0 0

v1v2v3

1− i 0 1− i0 −i 10 0 0

v1v2v3

Complex Eigenvalues

3− (2 + i) 1 10 2− (2 + i) 10 −1 2− (2 + i)

v1v2v3

1− i 1 10 −i 10 −1 −i

v1v2v3

1− i 1 10 −i 10 0 0

v1v2v3

1− i 0 1− i0 −i 10 0 0

v1v2v3

Complex Eigenvalues

3− (2 + i) 1 10 2− (2 + i) 10 −1 2− (2 + i)

v1v2v3

1− i 1 10 −i 10 −1 −i

v1v2v3

1− i 1 10 −i 10 0 0

v1v2v3

1− i 0 1− i0 −i 10 0 0

v1v2v3

Complex Eigenvalues

3− (2 + i) 1 10 2− (2 + i) 10 −1 2− (2 + i)

v1v2v3

1− i 1 10 −i 10 −1 −i

v1v2v3

1− i 1 10 −i 10 0 0

v1v2v3

1− i 0 1− i0 −i 10 0 0

v1v2v3

Complex Eigenvalues

3− (2 + i) 1 10 2− (2 + i) 10 −1 2− (2 + i)

v1v2v3

1− i 1 10 −i 10 −1 −i

v1v2v3

1− i 1 10 −i 10 0 0

v1v2v3

1− i 0 1− i0 −i 10 0 0

v1v2v3

Complex Eigenvalues

v(2) =

10−1

The corresponding solutions of the differential equation are

x(1) =

e3t ; x(2) = e2t

10−1

cos(t)−

sin(t)

x(3) = e2t

10−1

cos(t) +

sin(t)

Complex Eigenvalues

v(2) =

10−1

x(1) =

e3t ; x(2) = e2t

10−1

cos(t)−

sin(t)

x(3) = e2t

10−1

cos(t) +

sin(t)

Complex Eigenvalues

and a corresponding

eigenvector is

v(2) =

10−1

x(1) =

e3t ; x(2) = e2t

10−1

cos(t)−

sin(t)

x(3) = e2t

10−1

cos(t) +

sin(t)

Complex Eigenvalues

v(2) =

10−1

x(1) =

e3t ; x(2) = e2t

10−1

cos(t)−

sin(t)

x(3) = e2t

10−1

cos(t) +

sin(t)

Complex Eigenvalues

v(2) =

10−1

x(1) =

e3t ; x(2) = e2t

10−1

cos(t)−

sin(t)

x(3) = e2t

10−1

cos(t) +

sin(t)

Complex Eigenvalues

v(2) =

10−1

x(1) =

e3t ; x(2) = e2t

10−1

cos(t)−

sin(t)

x(3) = e2t

10−1

cos(t) +

sin(t)

Complex Eigenvalues

v(2) =

10−1

The corresponding

solutions of the differential equation are

x(1) =

e3t ; x(2) = e2t

10−1

cos(t)−

sin(t)

x(3) = e2t

10−1

cos(t) +

sin(t)

Complex Eigenvalues

v(2) =

10−1

The corresponding solutions

of the differential equation are

x(1) =

e3t ; x(2) = e2t

10−1

cos(t)−

sin(t)

x(3) = e2t

10−1

cos(t) +

sin(t)

Complex Eigenvalues

v(2) =

10−1

The corresponding solutions of the differential equation

x(1) =

e3t ; x(2) = e2t

10−1

cos(t)−

sin(t)

x(3) = e2t

10−1

cos(t) +

sin(t)

Complex Eigenvalues

v(2) =

10−1

x(1) =

e3t ; x(2) = e2t

10−1

cos(t)−

sin(t)

x(3) = e2t

10−1

cos(t) +

sin(t)

Complex Eigenvalues

v(2) =

10−1

x(1) =

x(2) = e2t

10−1

cos(t)−

sin(t)

x(3) = e2t

10−1

cos(t) +

sin(t)

Complex Eigenvalues

v(2) =

10−1

x(1) =

e3t ; x(2) = e2t

10−1

cos(t)−

sin(t)

x(3) = e2t

10−1

cos(t) +

sin(t)

Complex Eigenvalues

v(2) =

10−1

x(1) =

e3t ; x(2) = e2t

10−1

cos(t)−

sin(t)

x(3) = e2t

10−1

cos(t) +

sin(t)

Complex Eigenvalues

v(2) =

10−1

x(1) =

e3t ; x(2) = e2t

10−1

cos(t)−

sin(t)

x(3) = e2t

10−1

cos(t) +

sin(t)

Complex Eigenvalues

v(2) =

10−1

x(1) =

e3t ; x(2) = e2t

10−1

cos(t)−

sin(t)

x(3) = e2t

10−1

cos(t) +

sin(t)

Complex Eigenvalues

The Wronskian of these solutions is

W [x(1), x(2), x(3)](t) =

∣∣∣∣∣∣e3t e2tcos(t) e2tsin(t)0 −e2tsin(t) e2tcos(t)0 −e2tcos(t) −e2tsin(t)

∣∣∣∣∣∣ =

e3te2te2t

∣∣∣∣∣∣1 cos(t) sin(t)0 −sin(t) cos(t)0 −cos(t) −sin(t)

∣∣∣∣∣∣ =

e3te2te2t(sin2(t) + cos2(t)

)= e7t 6= 0

Complex Eigenvalues

The Wronskian

of these solutions is

W [x(1), x(2), x(3)](t) =

∣∣∣∣∣∣ =

e3te2te2t

∣∣∣∣∣∣ =

)= e7t 6= 0

Complex Eigenvalues

The Wronskian of these solutions

W [x(1), x(2), x(3)](t) =

∣∣∣∣∣∣ =

e3te2te2t

∣∣∣∣∣∣ =

)= e7t 6= 0

Complex Eigenvalues

W [x(1), x(2), x(3)](t) =

∣∣∣∣∣∣ =

e3te2te2t

∣∣∣∣∣∣ =

)= e7t 6= 0

Complex Eigenvalues

W [x(1), x(2), x(3)](t) =

∣∣∣∣∣∣ =

e3te2te2t

∣∣∣∣∣∣ =

)= e7t 6= 0

Complex Eigenvalues

W [x(1), x(2), x(3)](t) =

∣∣∣∣∣∣ =

e3te2te2t

∣∣∣∣∣∣ =

)= e7t 6= 0

Complex Eigenvalues

W [x(1), x(2), x(3)](t) =

∣∣∣∣∣∣ =

e3te2te2t

∣∣∣∣∣∣ =

)= e7t 6= 0

Complex Eigenvalues

W [x(1), x(2), x(3)](t) =

∣∣∣∣∣∣ =

e3te2te2t

∣∣∣∣∣∣ =

e7t 6= 0

Complex Eigenvalues

W [x(1), x(2), x(3)](t) =

∣∣∣∣∣∣ =

e3te2te2t

∣∣∣∣∣∣ =

)= e7t

Complex Eigenvalues

W [x(1), x(2), x(3)](t) =

∣∣∣∣∣∣ =

e3te2te2t

∣∣∣∣∣∣ =

)= e7t 6= 0

Complex Eigenvalues

Hence, the solutions x(1), x(2) and x(3) form a fundamental set,and the general solution of the system is

X = c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

e3t + c2

10−1

e2tcos(t)−

e2tsin(t)

10−1

e2tcos(t) +

e2tsin(t)

Complex Eigenvalues

Hence,

the solutions x(1), x(2) and x(3) form a fundamental set,and the general solution of the system is

X = c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

e3t + c2

10−1

e2tcos(t)−

e2tsin(t)

10−1

e2tcos(t) +

e2tsin(t)

Complex Eigenvalues

Hence, the solutions

x(1), x(2) and x(3) form a fundamental set,and the general solution of the system is

X = c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

e3t + c2

10−1

e2tcos(t)−

e2tsin(t)

10−1

e2tcos(t) +

e2tsin(t)

Complex Eigenvalues

Hence, the solutions x(1), x(2) and

x(3) form a fundamental set,and the general solution of the system is

X = c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

e3t + c2

10−1

e2tcos(t)−

e2tsin(t)

10−1

e2tcos(t) +

e2tsin(t)

Complex Eigenvalues

Hence, the solutions x(1), x(2) and x(3)

form a fundamental set,and the general solution of the system is

X = c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

e3t + c2

10−1

e2tcos(t)−

e2tsin(t)

10−1

e2tcos(t) +

e2tsin(t)

Complex Eigenvalues

Hence, the solutions x(1), x(2) and x(3) form a fundamental set,and

the general solution of the system is

X = c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

e3t + c2

10−1

e2tcos(t)−

e2tsin(t)

10−1

e2tcos(t) +

e2tsin(t)

Complex Eigenvalues

Hence, the solutions x(1), x(2) and x(3) form a fundamental set,and the general solution

of the system is

X = c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

e3t + c2

10−1

e2tcos(t)−

e2tsin(t)

10−1

e2tcos(t) +

e2tsin(t)

Complex Eigenvalues

Hence, the solutions x(1), x(2) and x(3) form a fundamental set,and the general solution of the system

X = c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

e3t + c2

10−1

e2tcos(t)−

e2tsin(t)

10−1

e2tcos(t) +

e2tsin(t)

Complex Eigenvalues

X = c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

e3t + c2

10−1

e2tcos(t)−

e2tsin(t)

10−1

e2tcos(t) +

e2tsin(t)

Complex Eigenvalues

c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

e3t + c2

10−1

e2tcos(t)−

e2tsin(t)

10−1

e2tcos(t) +

e2tsin(t)

Complex Eigenvalues

X = c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

e3t + c2

10−1

e2tcos(t)−

e2tsin(t)

10−1

e2tcos(t) +

e2tsin(t)

Complex Eigenvalues

X = c1x(1) + c2x(2) + c3x(3) =⇒

e3t + c2

10−1

e2tcos(t)−

e2tsin(t)

10−1

e2tcos(t) +

e2tsin(t)

Complex Eigenvalues

X = c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

10−1

e2tcos(t)−

e2tsin(t)

10−1

e2tcos(t) +

e2tsin(t)

Complex Eigenvalues

X = c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

e3t + c2

10−1

e2tcos(t)−

e2tsin(t)

10−1

e2tcos(t) +

e2tsin(t)

Complex Eigenvalues

X = c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

e3t + c2

10−1

e2tcos(t)−

e2tsin(t)

10−1

e2tcos(t) +

e2tsin(t)

Complex Eigenvalues

X = c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

e3t + c2

10−1

e2tcos(t)−

e2tsin(t)

10−1

e2tcos(t) +

e2tsin(t)

Complex Eigenvalues

X = c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

e3t + c2

10−1

e2tcos(t)−

e2tsin(t)

10−1

e2tcos(t) +

e2tsin(t)

Complex Eigenvalues

x1x2x3

c1e3t + e2t(c2cos(t) + c3sin(t))

0 e2t(−c2sin(t) + c3cos(t))0 −e2t(c2cos(t) + c3sin(t))

Here is the direction field associated with the system

x ′1x ′2x ′3

3 1 10 2 10 −1 2

x1x2x3

Complex Eigenvalues

x1x2x3

x ′1x ′2x ′3

3 1 10 2 10 −1 2

x1x2x3

Complex Eigenvalues

x1x2x3

x ′1x ′2x ′3

3 1 10 2 10 −1 2

x1x2x3

Complex Eigenvalues

x1x2x3

x ′1x ′2x ′3

3 1 10 2 10 −1 2

x1x2x3

Complex Eigenvalues

x1x2x3

Here is

the direction field associated with the system

x ′1x ′2x ′3

3 1 10 2 10 −1 2

x1x2x3

Complex Eigenvalues

x1x2x3

Here is the direction field

associated with the system

x ′1x ′2x ′3

3 1 10 2 10 −1 2

x1x2x3

Complex Eigenvalues

x1x2x3

Here is the direction field associated with

the system

x ′1x ′2x ′3

3 1 10 2 10 −1 2

x1x2x3

Complex Eigenvalues

x1x2x3

x ′1x ′2x ′3

3 1 10 2 10 −1 2

x1x2x3

Complex Eigenvalues

x1x2x3

x ′1x ′2x ′3

3 1 10 2 10 −1 2

x1x2x3

Complex Eigenvalues

x1x2x3

x ′1x ′2x ′3

3 1 10 2 10 −1 2

x1x2x3

Complex Eigenvalues

Example 9.7

X′ = AX =

(−1/2 1−1 −1/2

Solution

|A− λI| =

∣∣∣∣−1/2− λ 1−1 −1/2− λ

∣∣∣∣ = 0

(−1/2− λ)2 + 1 = (λ)2 + λ+5

Complex Eigenvalues

Example 9.7

X′ = AX =

(−1/2 1−1 −1/2

Solution

|A− λI| =

∣∣∣∣−1/2− λ 1−1 −1/2− λ

∣∣∣∣ = 0

(−1/2− λ)2 + 1 = (λ)2 + λ+5

Complex Eigenvalues

Example 9.7

the following ODE

X′ = AX =

(−1/2 1−1 −1/2

Solution

|A− λI| =

∣∣∣∣−1/2− λ 1−1 −1/2− λ

∣∣∣∣ = 0

(−1/2− λ)2 + 1 = (λ)2 + λ+5

Complex Eigenvalues

Example 9.7

X′ = AX =

(−1/2 1−1 −1/2

Solution

|A− λI| =

∣∣∣∣−1/2− λ 1−1 −1/2− λ

∣∣∣∣ = 0

(−1/2− λ)2 + 1 = (λ)2 + λ+5

Complex Eigenvalues

Example 9.7

X′ = AX =

(−1/2 1−1 −1/2

Solution

|A− λI| =

∣∣∣∣−1/2− λ 1−1 −1/2− λ

∣∣∣∣ = 0

(−1/2− λ)2 + 1 = (λ)2 + λ+5

Complex Eigenvalues

Example 9.7

X′ = AX =

(−1/2 1−1 −1/2

Solution

|A− λI| =

∣∣∣∣−1/2− λ 1−1 −1/2− λ

∣∣∣∣ = 0

(−1/2− λ)2 + 1 = (λ)2 + λ+5

Complex Eigenvalues

Example 9.7

X′ = AX =

(−1/2 1−1 −1/2

Solution

|A− λI| =

∣∣∣∣−1/2− λ 1−1 −1/2− λ

∣∣∣∣ = 0

(−1/2− λ)2 + 1 = (λ)2 + λ+5

Complex Eigenvalues

Example 9.7

X′ = AX =

(−1/2 1−1 −1/2

Solution

Let’s find

|A− λI| =

∣∣∣∣−1/2− λ 1−1 −1/2− λ

∣∣∣∣ = 0

(−1/2− λ)2 + 1 = (λ)2 + λ+5

Complex Eigenvalues

Example 9.7

X′ = AX =

(−1/2 1−1 −1/2

Solution

of the matrix A

|A− λI| =

∣∣∣∣−1/2− λ 1−1 −1/2− λ

∣∣∣∣ = 0

(−1/2− λ)2 + 1 = (λ)2 + λ+5

Complex Eigenvalues

Example 9.7

X′ = AX =

(−1/2 1−1 −1/2

Solution

|A− λI| =

∣∣∣∣−1/2− λ 1−1 −1/2− λ

∣∣∣∣ = 0

(−1/2− λ)2 + 1 = (λ)2 + λ+5

Complex Eigenvalues

Example 9.7

X′ = AX =

(−1/2 1−1 −1/2

Solution

|A− λI| =

∣∣∣∣−1/2− λ 1−1 −1/2− λ

∣∣∣∣ = 0

(−1/2− λ)2 + 1 = (λ)2 + λ+5

Complex Eigenvalues

Example 9.7

X′ = AX =

(−1/2 1−1 −1/2

Solution

|A− λI| =

∣∣∣∣−1/2− λ 1−1 −1/2− λ

∣∣∣∣ = 0

(−1/2− λ)2 + 1 = (λ)2 + λ+5

Complex Eigenvalues

Example 9.7

X′ = AX =

(−1/2 1−1 −1/2

Solution

|A− λI| =

∣∣∣∣−1/2− λ 1−1 −1/2− λ

∣∣∣∣ = 0

(−1/2− λ)2 + 1 = (λ)2 + λ+5

Complex Eigenvalues

Example 9.7

X′ = AX =

(−1/2 1−1 −1/2

Solution

|A− λI| =

∣∣∣∣−1/2− λ 1−1 −1/2− λ

∣∣∣∣ = 0

(−1/2− λ)2 + 1 =

(λ)2 + λ+5

Complex Eigenvalues

Example 9.7

X′ = AX =

(−1/2 1−1 −1/2

Solution

|A− λI| =

∣∣∣∣−1/2− λ 1−1 −1/2− λ

∣∣∣∣ = 0

(−1/2− λ)2 + 1 = (λ)2 + λ+5

Complex Eigenvalues

λ1 = −1

2+ i , λ2 = −1

2− i

If λ1 = −12 + i , then

(A− λ1I) x =

(−1/2− λ 1−1 −1/2− λ

)(v1v2

(−1/2− (−1

2 + i) 1−1 −1/2− (−1

2 + i)

)(v1v2

−i 1−1 −i

)(v1v2

Complex Eigenvalues

λ1 = −1

2+ i ,

λ2 = −1

2− i

If λ1 = −12 + i , then

(A− λ1I) x =

(−1/2− λ 1−1 −1/2− λ

)(v1v2

(−1/2− (−1

2 + i) 1−1 −1/2− (−1

2 + i)

)(v1v2

−i 1−1 −i

)(v1v2

Complex Eigenvalues

λ1 = −1

2+ i , λ2 = −1

2− i

If λ1 = −12 + i , then

(A− λ1I) x =

(−1/2− λ 1−1 −1/2− λ

)(v1v2

(−1/2− (−1

2 + i) 1−1 −1/2− (−1

2 + i)

)(v1v2

−i 1−1 −i

)(v1v2

Complex Eigenvalues

λ1 = −1

2+ i , λ2 = −1

2− i

λ1 = −12 + i , then

(A− λ1I) x =

(−1/2− λ 1−1 −1/2− λ

)(v1v2

(−1/2− (−1

2 + i) 1−1 −1/2− (−1

2 + i)

)(v1v2

−i 1−1 −i

)(v1v2

Complex Eigenvalues

λ1 = −1

2+ i , λ2 = −1

2− i

If λ1 = −12 + i ,

(A− λ1I) x =

(−1/2− λ 1−1 −1/2− λ

)(v1v2

(−1/2− (−1

2 + i) 1−1 −1/2− (−1

2 + i)

)(v1v2

−i 1−1 −i

)(v1v2

Complex Eigenvalues

λ1 = −1

2+ i , λ2 = −1

2− i

If λ1 = −12 + i , then

(A− λ1I) x =

(−1/2− λ 1−1 −1/2− λ

)(v1v2

(−1/2− (−1

2 + i) 1−1 −1/2− (−1

2 + i)

)(v1v2

−i 1−1 −i

)(v1v2

Complex Eigenvalues

λ1 = −1

2+ i , λ2 = −1

2− i

If λ1 = −12 + i , then

(A− λ1I) x =

(−1/2− λ 1−1 −1/2− λ

)(v1v2

(−1/2− (−1

2 + i) 1−1 −1/2− (−1

2 + i)

)(v1v2

−i 1−1 −i

)(v1v2

Complex Eigenvalues

λ1 = −1

2+ i , λ2 = −1

2− i

If λ1 = −12 + i , then

(A− λ1I) x =

(−1/2− λ 1−1 −1/2− λ

)(v1v2

(−1/2− (−1

2 + i) 1−1 −1/2− (−1

2 + i)

)(v1v2

−i 1−1 −i

)(v1v2

Complex Eigenvalues

λ1 = −1

2+ i , λ2 = −1

2− i

If λ1 = −12 + i , then

(A− λ1I) x =

(−1/2− λ 1−1 −1/2− λ

)(v1v2

(−1/2− (−1

2 + i) 1−1 −1/2− (−1

2 + i)

)(v1v2

(−i 1−1 −i

)(v1v2

Complex Eigenvalues

λ1 = −1

2+ i , λ2 = −1

2− i

If λ1 = −12 + i , then

(A− λ1I) x =

(−1/2− λ 1−1 −1/2− λ

)(v1v2

(−1/2− (−1

2 + i) 1−1 −1/2− (−1

2 + i)

)(v1v2

−i 1−1 −i

)(v1v2

Complex Eigenvalues

v(1) =

)If λ2 = −1

2 − i , then

(A− λ1I) x =

(−1/2− (−1

2 − i) 1−1 −1/2− (−1

2 − i)

)(v1v2

Complex Eigenvalues

v(1) =

)If λ2 = −1

2 − i , then

(A− λ1I) x =

(−1/2− (−1

2 − i) 1−1 −1/2− (−1

2 − i)

)(v1v2

Complex Eigenvalues

and a corresponding

eigenvector is

v(1) =

)If λ2 = −1

2 − i , then

(A− λ1I) x =

(−1/2− (−1

2 − i) 1−1 −1/2− (−1

2 − i)

)(v1v2

Complex Eigenvalues

v(1) =

)If λ2 = −1

2 − i , then

(A− λ1I) x =

(−1/2− (−1

2 − i) 1−1 −1/2− (−1

2 − i)

)(v1v2

Complex Eigenvalues

v(1) =

If λ2 = −12 − i , then

(A− λ1I) x =

(−1/2− (−1

2 − i) 1−1 −1/2− (−1

2 − i)

)(v1v2

Complex Eigenvalues

v(1) =

λ2 = −12 − i , then

(A− λ1I) x =

(−1/2− (−1

2 − i) 1−1 −1/2− (−1

2 − i)

)(v1v2

Complex Eigenvalues

v(1) =

)If λ2 = −1

2 − i ,

(A− λ1I) x =

(−1/2− (−1

2 − i) 1−1 −1/2− (−1

2 − i)

)(v1v2

Complex Eigenvalues

v(1) =

)If λ2 = −1

2 − i , then

(A− λ1I) x =

(−1/2− (−1

2 − i) 1−1 −1/2− (−1

2 − i)

)(v1v2

Complex Eigenvalues

v(1) =

)If λ2 = −1

2 − i , then

(A− λ1I) x =

(−1/2− (−1

2 − i) 1−1 −1/2− (−1

2 − i)

)(v1v2

Complex Eigenvalues

v(1) =

)If λ2 = −1

2 − i , then

(A− λ1I) x =

(−1/2− (−1

2 − i) 1−1 −1/2− (−1

2 − i)

)(v1v2

Complex Eigenvalues

(i 1−1 i

)(v1v2

v(2) =

(1−i

)The corresponding solutions of the differential equation are

x(1) =

)e(−1/2+i)t ; x(2) =

)e(−1/2−i)t

Complex Eigenvalues

(i 1−1 i

)(v1v2

v(2) =

(1−i

x(1) =

)e(−1/2+i)t ; x(2) =

)e(−1/2−i)t

Complex Eigenvalues

(i 1−1 i

)(v1v2

v(2) =

(1−i

x(1) =

)e(−1/2+i)t ; x(2) =

)e(−1/2−i)t

Complex Eigenvalues

(i 1−1 i

)(v1v2

and a corresponding

eigenvector is

v(2) =

(1−i

x(1) =

)e(−1/2+i)t ; x(2) =

)e(−1/2−i)t

Complex Eigenvalues

(i 1−1 i

)(v1v2

v(2) =

(1−i

x(1) =

)e(−1/2+i)t ; x(2) =

)e(−1/2−i)t

Complex Eigenvalues

(i 1−1 i

)(v1v2

v(2) =

(1−i

x(1) =

)e(−1/2+i)t ; x(2) =

)e(−1/2−i)t

Complex Eigenvalues

(i 1−1 i

)(v1v2

v(2) =

(1−i

)The corresponding solutions

of the differential equation are

x(1) =

)e(−1/2+i)t ; x(2) =

)e(−1/2−i)t

Complex Eigenvalues

(i 1−1 i

)(v1v2

v(2) =

(1−i

)The corresponding solutions of the differential equation

x(1) =

)e(−1/2+i)t ; x(2) =

)e(−1/2−i)t

Complex Eigenvalues

(i 1−1 i

)(v1v2

v(2) =

(1−i

x(1) =

)e(−1/2+i)t ; x(2) =

)e(−1/2−i)t

Complex Eigenvalues

(i 1−1 i

)(v1v2

v(2) =

(1−i

x(1) =

)e(−1/2+i)t ;

x(2) =

)e(−1/2−i)t

Complex Eigenvalues

(i 1−1 i

)(v1v2

v(2) =

(1−i

x(1) =

)e(−1/2+i)t ; x(2) =

)e(−1/2−i)t

Complex Eigenvalues

(i 1−1 i

)(v1v2

v(2) =

(1−i

x(1) =

)e(−1/2+i)t ; x(2) =

)e(−1/2−i)t

Complex Eigenvalues

To obtain a set of real-valued solutions, we can choose the realand imaginary parts of either x (1) or x (2). In fact,

x(1) =

)e(−1/2+i)t =

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

(e−t/2sin(t)

e−t/2cos(t)

)Hence, a set of real-valued solutions are

u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

Complex Eigenvalues

To obtain

a set of real-valued solutions, we can choose the realand imaginary parts of either x (1) or x (2). In fact,

x(1) =

)e(−1/2+i)t =

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

(e−t/2sin(t)

e−t/2cos(t)

u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

Complex Eigenvalues

To obtain a set of

real-valued solutions, we can choose the realand imaginary parts of either x (1) or x (2). In fact,

x(1) =

)e(−1/2+i)t =

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

(e−t/2sin(t)

e−t/2cos(t)

u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

Complex Eigenvalues

To obtain a set of real-valued solutions,

we can choose the realand imaginary parts of either x (1) or x (2). In fact,

x(1) =

)e(−1/2+i)t =

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

(e−t/2sin(t)

e−t/2cos(t)

u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

Complex Eigenvalues

To obtain a set of real-valued solutions, we can choose

the realand imaginary parts of either x (1) or x (2). In fact,

x(1) =

)e(−1/2+i)t =

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

(e−t/2sin(t)

e−t/2cos(t)

u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

Complex Eigenvalues

To obtain a set of real-valued solutions, we can choose the realand

imaginary parts of either x (1) or x (2). In fact,

x(1) =

)e(−1/2+i)t =

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

(e−t/2sin(t)

e−t/2cos(t)

u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

Complex Eigenvalues

To obtain a set of real-valued solutions, we can choose the realand imaginary parts

of either x (1) or x (2). In fact,

x(1) =

)e(−1/2+i)t =

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

(e−t/2sin(t)

e−t/2cos(t)

u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

Complex Eigenvalues

To obtain a set of real-valued solutions, we can choose the realand imaginary parts of either x (1) or

x (2). In fact,

x(1) =

)e(−1/2+i)t =

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

(e−t/2sin(t)

e−t/2cos(t)

u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

Complex Eigenvalues

To obtain a set of real-valued solutions, we can choose the realand imaginary parts of either x (1) or x (2).

In fact,

x(1) =

)e(−1/2+i)t =

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

(e−t/2sin(t)

e−t/2cos(t)

u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

Complex Eigenvalues

x(1) =

)e(−1/2+i)t =

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

(e−t/2sin(t)

e−t/2cos(t)

u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

Complex Eigenvalues

x(1) =

)e(−1/2+i)t =

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

(e−t/2sin(t)

e−t/2cos(t)

u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

Complex Eigenvalues

x(1) =

)e(−1/2+i)t =

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

(e−t/2sin(t)

e−t/2cos(t)

u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

Complex Eigenvalues

x(1) =

)e(−1/2+i)t =

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

(e−t/2sin(t)

e−t/2cos(t)

u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

Complex Eigenvalues

x(1) =

)e(−1/2+i)t =

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

(e−t/2sin(t)

e−t/2cos(t)

Hence, a set of real-valued solutions are

u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

Complex Eigenvalues

x(1) =

)e(−1/2+i)t =

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

(e−t/2sin(t)

e−t/2cos(t)

)Hence,

a set of real-valued solutions are

u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

Complex Eigenvalues

x(1) =

)e(−1/2+i)t =

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

(e−t/2sin(t)

e−t/2cos(t)

)Hence, a set of

real-valued solutions are

u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

Complex Eigenvalues

x(1) =

)e(−1/2+i)t =

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

(e−t/2sin(t)

e−t/2cos(t)

)Hence, a set of real-valued

solutions are

u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

Complex Eigenvalues

x(1) =

)e(−1/2+i)t =

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

(e−t/2sin(t)

e−t/2cos(t)

)Hence, a set of real-valued solutions

u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

Complex Eigenvalues

x(1) =

)e(−1/2+i)t =

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

(e−t/2sin(t)

e−t/2cos(t)

u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

Complex Eigenvalues

x(1) =

)e(−1/2+i)t =

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

(e−t/2sin(t)

e−t/2cos(t)

u(t) = e−t/2(

cos(t)−sin(t)

v(t) = e−t/2(sin(t)cos(t)

Complex Eigenvalues

x(1) =

)e(−1/2+i)t =

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

(e−t/2sin(t)

e−t/2cos(t)

u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

Complex Eigenvalues

The Wronskian of these two real-valued solutions is

W [x(1), x(2)](t) =

∣∣∣∣ e−t/2cos(t) e−t/2sin(t)

−e−t/2sin(t) e−t/2cos(t)

∣∣∣∣ =

e−t/2e−t/2∣∣∣∣ cos(t) sin(t)−sin(t) cos(t)

∣∣∣∣ = e−t 6= 0

Hence, the solutions x(1), x(2) form a fundamental set,

Complex Eigenvalues

The Wronskian

of these two real-valued solutions is

W [x(1), x(2)](t) =

∣∣∣∣ =

∣∣∣∣ = e−t 6= 0

Complex Eigenvalues

The Wronskian of these two

real-valued solutions is

W [x(1), x(2)](t) =

∣∣∣∣ =

∣∣∣∣ = e−t 6= 0

Complex Eigenvalues

The Wronskian of these two real-valued solutions

W [x(1), x(2)](t) =

∣∣∣∣ =

∣∣∣∣ = e−t 6= 0

Complex Eigenvalues

W [x(1), x(2)](t) =

∣∣∣∣ =

∣∣∣∣ = e−t 6= 0

Complex Eigenvalues

W [x(1), x(2)](t) =

∣∣∣∣ =

∣∣∣∣ = e−t 6= 0

Complex Eigenvalues

W [x(1), x(2)](t) =

∣∣∣∣ =

∣∣∣∣ = e−t 6= 0

Complex Eigenvalues

W [x(1), x(2)](t) =

∣∣∣∣ =

∣∣∣∣ = e−t

Complex Eigenvalues

W [x(1), x(2)](t) =

∣∣∣∣ =

∣∣∣∣ = e−t 6= 0

Complex Eigenvalues

W [x(1), x(2)](t) =

∣∣∣∣ =

∣∣∣∣ = e−t 6= 0

Hence,

the solutions x(1), x(2) form a fundamental set,

Complex Eigenvalues

W [x(1), x(2)](t) =

∣∣∣∣ =

∣∣∣∣ = e−t 6= 0

Hence, the solutions

x(1), x(2) form a fundamental set,

Complex Eigenvalues

W [x(1), x(2)](t) =

∣∣∣∣ =

∣∣∣∣ = e−t 6= 0

Hence, the solutions x(1),

x(2) form a fundamental set,

Complex Eigenvalues

W [x(1), x(2)](t) =

∣∣∣∣ =

∣∣∣∣ = e−t 6= 0

Hence, the solutions x(1), x(2)

form a fundamental set,

Complex Eigenvalues

W [x(1), x(2)](t) =

∣∣∣∣ =

∣∣∣∣ = e−t 6= 0

Hence, the solutions x(1), x(2) form a

fundamental set,

Complex Eigenvalues

W [x(1), x(2)](t) =

∣∣∣∣ =

∣∣∣∣ = e−t 6= 0

Complex Eigenvalues

and the general solution of the system is

X = c1x(1) + c2x(2) = c1e−t/2

(cos(t)−sin(t)

)+ c2e

−t/2(sin(t)cos(t)

)Here is the direction field associated with the system(

x ′1x ′2

(−1/2 1−1 −1/2

)(x1x2

Complex Eigenvalues

the general solution of the system is

X = c1x(1) + c2x(2) = c1e−t/2

(cos(t)−sin(t)

)+ c2e

−t/2(sin(t)cos(t)

x ′1x ′2

(−1/2 1−1 −1/2

)(x1x2

Complex Eigenvalues

and the general solution

of the system is

X = c1x(1) + c2x(2) = c1e−t/2

(cos(t)−sin(t)

)+ c2e

−t/2(sin(t)cos(t)

x ′1x ′2

(−1/2 1−1 −1/2

)(x1x2

Complex Eigenvalues

and the general solution of the system

X = c1x(1) + c2x(2) = c1e−t/2

(cos(t)−sin(t)

)+ c2e

−t/2(sin(t)cos(t)

x ′1x ′2

(−1/2 1−1 −1/2

)(x1x2

Complex Eigenvalues

X = c1x(1) + c2x(2) = c1e−t/2

(cos(t)−sin(t)

)+ c2e

−t/2(sin(t)cos(t)

x ′1x ′2

(−1/2 1−1 −1/2

)(x1x2

Complex Eigenvalues

c1x(1) + c2x(2) = c1e−t/2

(cos(t)−sin(t)

)+ c2e

−t/2(sin(t)cos(t)

x ′1x ′2

(−1/2 1−1 −1/2

)(x1x2

Complex Eigenvalues

X = c1x(1) + c2x(2) =

c1e−t/2

(cos(t)−sin(t)

)+ c2e

−t/2(sin(t)cos(t)

x ′1x ′2

(−1/2 1−1 −1/2

)(x1x2

Complex Eigenvalues

X = c1x(1) + c2x(2) = c1e−t/2

(cos(t)−sin(t)

c2e−t/2

(sin(t)cos(t)

x ′1x ′2

(−1/2 1−1 −1/2

)(x1x2

Complex Eigenvalues

X = c1x(1) + c2x(2) = c1e−t/2

(cos(t)−sin(t)

)+ c2e

−t/2(sin(t)cos(t)

Here is the direction field associated with the system(x ′1x ′2

(−1/2 1−1 −1/2

)(x1x2

Complex Eigenvalues

X = c1x(1) + c2x(2) = c1e−t/2

(cos(t)−sin(t)

)+ c2e

−t/2(sin(t)cos(t)

)Here is

the direction field associated with the system(x ′1x ′2

(−1/2 1−1 −1/2

)(x1x2

Complex Eigenvalues

X = c1x(1) + c2x(2) = c1e−t/2

(cos(t)−sin(t)

)+ c2e

−t/2(sin(t)cos(t)

)Here is the direction field

associated with the system(x ′1x ′2

(−1/2 1−1 −1/2

)(x1x2

Complex Eigenvalues

X = c1x(1) + c2x(2) = c1e−t/2

(cos(t)−sin(t)

)+ c2e

−t/2(sin(t)cos(t)

)Here is the direction field associated with

the system(x ′1x ′2

(−1/2 1−1 −1/2

)(x1x2

Complex Eigenvalues

X = c1x(1) + c2x(2) = c1e−t/2

(cos(t)−sin(t)

)+ c2e

−t/2(sin(t)cos(t)

)Here is the direction field associated with the system

(x ′1x ′2

(−1/2 1−1 −1/2

)(x1x2

Complex Eigenvalues

X = c1x(1) + c2x(2) = c1e−t/2

(cos(t)−sin(t)

)+ c2e

−t/2(sin(t)cos(t)

x ′1x ′2

(−1/2 1−1 −1/2

)(x1x2

Complex Eigenvalues

X = c1x(1) + c2x(2) = c1e−t/2

(cos(t)−sin(t)

)+ c2e

−t/2(sin(t)cos(t)

x ′1x ′2

(−1/2 1−1 −1/2

)(x1x2

Complex Eigenvalues

Repeated Eigenvalues

We conclude our consideration of the linear homogeneous systemwith constant coefficients

x′ = Ax

with a discussion of the case in which the matrix A has a repeatedeigenvalues. suppose that λ is a repetead root of the characteristicequation ∣∣∣A− λI

∣∣∣ = 0

Then, λ is an eigenvalue of algebraic multiplicity 2 of the matrixA. In this event, there are two possibilities: The matrx A isnon-defective and there is still a fundamental set of solutions ofthe form

{veλt ,weλt

We conclude

our consideration of the linear homogeneous systemwith constant coefficients

x′ = Ax

∣∣∣ = 0

{veλt ,weλt

We conclude our consideration

of the linear homogeneous systemwith constant coefficients

x′ = Ax

∣∣∣ = 0

{veλt ,weλt

We conclude our consideration of the linear homogeneous system

with constant coefficients

x′ = Ax

∣∣∣ = 0

{veλt ,weλt

x′ = Ax

∣∣∣ = 0

{veλt ,weλt

x′ = Ax

∣∣∣ = 0

{veλt ,weλt

x′ = Ax

with a discussion

of the case in which the matrix A has a repeatedeigenvalues. suppose that λ is a repetead root of the characteristicequation ∣∣∣A− λI

∣∣∣ = 0

{veλt ,weλt

x′ = Ax

with a discussion of the case

in which the matrix A has a repeatedeigenvalues. suppose that λ is a repetead root of the characteristicequation ∣∣∣A− λI

∣∣∣ = 0

{veλt ,weλt

x′ = Ax

with a discussion of the case in which the matrix

A has a repeatedeigenvalues. suppose that λ is a repetead root of the characteristicequation ∣∣∣A− λI

∣∣∣ = 0

{veλt ,weλt

x′ = Ax

with a discussion of the case in which the matrix A

has a repeatedeigenvalues. suppose that λ is a repetead root of the characteristicequation ∣∣∣A− λI

∣∣∣ = 0

{veλt ,weλt

x′ = Ax

with a discussion of the case in which the matrix A has a repeatedeigenvalues.

suppose that λ is a repetead root of the characteristicequation ∣∣∣A− λI

∣∣∣ = 0

{veλt ,weλt

x′ = Ax

with a discussion of the case in which the matrix A has a repeatedeigenvalues. suppose that

λ is a repetead root of the characteristicequation ∣∣∣A− λI

∣∣∣ = 0

{veλt ,weλt

x′ = Ax

with a discussion of the case in which the matrix A has a repeatedeigenvalues. suppose that λ

is a repetead root of the characteristicequation ∣∣∣A− λI

∣∣∣ = 0

{veλt ,weλt

x′ = Ax

with a discussion of the case in which the matrix A has a repeatedeigenvalues. suppose that λ is a repetead root

of the characteristicequation ∣∣∣A− λI

∣∣∣ = 0

{veλt ,weλt

x′ = Ax

with a discussion of the case in which the matrix A has a repeatedeigenvalues. suppose that λ is a repetead root of the

characteristicequation ∣∣∣A− λI

∣∣∣ = 0

{veλt ,weλt

x′ = Ax

with a discussion of the case in which the matrix A has a repeatedeigenvalues. suppose that λ is a repetead root of the characteristicequation

∣∣∣A− λI∣∣∣ = 0

{veλt ,weλt

x′ = Ax

∣∣∣ = 0

{veλt ,weλt

x′ = Ax

∣∣∣ = 0

λ is an eigenvalue of algebraic multiplicity 2 of the matrixA. In this event, there are two possibilities: The matrx A isnon-defective and there is still a fundamental set of solutions ofthe form

{veλt ,weλt

x′ = Ax

∣∣∣ = 0

Then, λ

is an eigenvalue of algebraic multiplicity 2 of the matrixA. In this event, there are two possibilities: The matrx A isnon-defective and there is still a fundamental set of solutions ofthe form

{veλt ,weλt

x′ = Ax

∣∣∣ = 0

Then, λ is an eigenvalue

of algebraic multiplicity 2 of the matrixA. In this event, there are two possibilities: The matrx A isnon-defective and there is still a fundamental set of solutions ofthe form

{veλt ,weλt

x′ = Ax

∣∣∣ = 0

Then, λ is an eigenvalue of algebraic multiplicity

2 of the matrixA. In this event, there are two possibilities: The matrx A isnon-defective and there is still a fundamental set of solutions ofthe form

{veλt ,weλt

x′ = Ax

∣∣∣ = 0

Then, λ is an eigenvalue of algebraic multiplicity 2

of the matrixA. In this event, there are two possibilities: The matrx A isnon-defective and there is still a fundamental set of solutions ofthe form

{veλt ,weλt

x′ = Ax

∣∣∣ = 0

Then, λ is an eigenvalue of algebraic multiplicity 2 of the matrixA.

In this event, there are two possibilities: The matrx A isnon-defective and there is still a fundamental set of solutions ofthe form

{veλt ,weλt

x′ = Ax

∣∣∣ = 0

Then, λ is an eigenvalue of algebraic multiplicity 2 of the matrixA. In this event,

there are two possibilities: The matrx A isnon-defective and there is still a fundamental set of solutions ofthe form

{veλt ,weλt

x′ = Ax

∣∣∣ = 0

Then, λ is an eigenvalue of algebraic multiplicity 2 of the matrixA. In this event, there are

two possibilities: The matrx A isnon-defective and there is still a fundamental set of solutions ofthe form

{veλt ,weλt

x′ = Ax

∣∣∣ = 0

Then, λ is an eigenvalue of algebraic multiplicity 2 of the matrixA. In this event, there are two possibilities:

The matrx A isnon-defective and there is still a fundamental set of solutions ofthe form

{veλt ,weλt

x′ = Ax

∣∣∣ = 0

Then, λ is an eigenvalue of algebraic multiplicity 2 of the matrixA. In this event, there are two possibilities: The matrx A

isnon-defective and there is still a fundamental set of solutions ofthe form

{veλt ,weλt

x′ = Ax

∣∣∣ = 0

Then, λ is an eigenvalue of algebraic multiplicity 2 of the matrixA. In this event, there are two possibilities: The matrx A isnon-defective and

there is still a fundamental set of solutions ofthe form

{veλt ,weλt

x′ = Ax

∣∣∣ = 0

Then, λ is an eigenvalue of algebraic multiplicity 2 of the matrixA. In this event, there are two possibilities: The matrx A isnon-defective and

there is still a fundamental set of solutions ofthe form

{veλt ,weλt

x′ = Ax

∣∣∣ = 0

Then, λ is an eigenvalue of algebraic multiplicity 2 of the matrixA. In this event, there are two possibilities: The matrx A isnon-defective and there is still

a fundamental set of solutions ofthe form

{veλt ,weλt

x′ = Ax

∣∣∣ = 0

Then, λ is an eigenvalue of algebraic multiplicity 2 of the matrixA. In this event, there are two possibilities: The matrx A isnon-defective and there is still a fundamental set of solutions

ofthe form

{veλt ,weλt

x′ = Ax

∣∣∣ = 0

{veλt ,weλt

x′ = Ax

∣∣∣ = 0

{veλt ,

weλt}

x′ = Ax

∣∣∣ = 0

{veλt ,weλt

However, if the matrx A is defective, there is just one solution ofthe form veλt associated with this eigenvalue. Therefore, toconstruct the general solution, it is necessary to find other solutionof a different form.

Recall that a similar situation occurred for the linear equationay ′′ + by ′ + cy = 0 when the characteristic equation had a doubleroot r . In that case we found one exponential solution y1(t) = ert ,but a second independent solution had the form y2(t) = tert

In this way, it may be natural to attempt to find a secondindependent solution of the form

x = wteλt

However,

if the matrx A is defective, there is just one solution ofthe form veλt associated with this eigenvalue. Therefore, toconstruct the general solution, it is necessary to find other solutionof a different form.

x = wteλt

However, if the matrx A

is defective, there is just one solution ofthe form veλt associated with this eigenvalue. Therefore, toconstruct the general solution, it is necessary to find other solutionof a different form.

x = wteλt

However, if the matrx A is defective,

there is just one solution ofthe form veλt associated with this eigenvalue. Therefore, toconstruct the general solution, it is necessary to find other solutionof a different form.

x = wteλt

However, if the matrx A is defective, there is

just one solution ofthe form veλt associated with this eigenvalue. Therefore, toconstruct the general solution, it is necessary to find other solutionof a different form.

x = wteλt

However, if the matrx A is defective, there is just one solution

ofthe form veλt associated with this eigenvalue. Therefore, toconstruct the general solution, it is necessary to find other solutionof a different form.

x = wteλt

However, if the matrx A is defective, there is just one solution ofthe form

veλt associated with this eigenvalue. Therefore, toconstruct the general solution, it is necessary to find other solutionof a different form.

x = wteλt

However, if the matrx A is defective, there is just one solution ofthe form veλt

associated with this eigenvalue. Therefore, toconstruct the general solution, it is necessary to find other solutionof a different form.

x = wteλt

However, if the matrx A is defective, there is just one solution ofthe form veλt associated with

this eigenvalue. Therefore, toconstruct the general solution, it is necessary to find other solutionof a different form.

x = wteλt

However, if the matrx A is defective, there is just one solution ofthe form veλt associated with this eigenvalue.

Therefore, toconstruct the general solution, it is necessary to find other solutionof a different form.

x = wteλt

However, if the matrx A is defective, there is just one solution ofthe form veλt associated with this eigenvalue. Therefore,

toconstruct the general solution, it is necessary to find other solutionof a different form.

x = wteλt

However, if the matrx A is defective, there is just one solution ofthe form veλt associated with this eigenvalue. Therefore, toconstruct the general solution,

it is necessary to find other solutionof a different form.

x = wteλt

However, if the matrx A is defective, there is just one solution ofthe form veλt associated with this eigenvalue. Therefore, toconstruct the general solution, it is necessary

to find other solutionof a different form.

x = wteλt

However, if the matrx A is defective, there is just one solution ofthe form veλt associated with this eigenvalue. Therefore, toconstruct the general solution, it is necessary to find

other solutionof a different form.

x = wteλt

However, if the matrx A is defective, there is just one solution ofthe form veλt associated with this eigenvalue. Therefore, toconstruct the general solution, it is necessary to find other solution

of a different form.

x = wteλt

Recall that

a similar situation occurred for the linear equationay ′′ + by ′ + cy = 0 when the characteristic equation had a doubleroot r . In that case we found one exponential solution y1(t) = ert ,but a second independent solution had the form y2(t) = tert

x = wteλt

Recall that a similar situation

occurred for the linear equationay ′′ + by ′ + cy = 0 when the characteristic equation had a doubleroot r . In that case we found one exponential solution y1(t) = ert ,but a second independent solution had the form y2(t) = tert

x = wteλt

Recall that a similar situation occurred for

the linear equationay ′′ + by ′ + cy = 0 when the characteristic equation had a doubleroot r . In that case we found one exponential solution y1(t) = ert ,but a second independent solution had the form y2(t) = tert

x = wteλt

Recall that a similar situation occurred for the linear equationay ′′ + by ′ + cy = 0

when the characteristic equation had a doubleroot r . In that case we found one exponential solution y1(t) = ert ,but a second independent solution had the form y2(t) = tert

x = wteλt

Recall that a similar situation occurred for the linear equationay ′′ + by ′ + cy = 0 when the characteristic equation

had a doubleroot r . In that case we found one exponential solution y1(t) = ert ,but a second independent solution had the form y2(t) = tert

x = wteλt

Recall that a similar situation occurred for the linear equationay ′′ + by ′ + cy = 0 when the characteristic equation had a doubleroot r .

In that case we found one exponential solution y1(t) = ert ,but a second independent solution had the form y2(t) = tert

x = wteλt

Recall that a similar situation occurred for the linear equationay ′′ + by ′ + cy = 0 when the characteristic equation had a doubleroot r . In that case

we found one exponential solution y1(t) = ert ,but a second independent solution had the form y2(t) = tert

x = wteλt

Recall that a similar situation occurred for the linear equationay ′′ + by ′ + cy = 0 when the characteristic equation had a doubleroot r . In that case we found

one exponential solution y1(t) = ert ,but a second independent solution had the form y2(t) = tert

x = wteλt

Recall that a similar situation occurred for the linear equationay ′′ + by ′ + cy = 0 when the characteristic equation had a doubleroot r . In that case we found one exponential solution

y1(t) = ert ,but a second independent solution had the form y2(t) = tert

x = wteλt

Recall that a similar situation occurred for the linear equationay ′′ + by ′ + cy = 0 when the characteristic equation had a doubleroot r . In that case we found one exponential solution y1(t) = ert ,

but a second independent solution had the form y2(t) = tert

x = wteλt

Recall that a similar situation occurred for the linear equationay ′′ + by ′ + cy = 0 when the characteristic equation had a doubleroot r . In that case we found one exponential solution y1(t) = ert ,but a second

independent solution had the form y2(t) = tert

x = wteλt

Recall that a similar situation occurred for the linear equationay ′′ + by ′ + cy = 0 when the characteristic equation had a doubleroot r . In that case we found one exponential solution y1(t) = ert ,but a second independent solution

had the form y2(t) = tert

x = wteλt

Recall that a similar situation occurred for the linear equationay ′′ + by ′ + cy = 0 when the characteristic equation had a doubleroot r . In that case we found one exponential solution y1(t) = ert ,but a second independent solution had the form

y2(t) = tert

x = wteλt

In this way,

it may be natural to attempt to find a secondindependent solution of the form

x = wteλt

In this way, it may be natural

to attempt to find a secondindependent solution of the form

x = wteλt

In this way, it may be natural to attempt to find

a secondindependent solution of the form

x = wteλt

In this way, it may be natural to attempt to find a second

independent solution of the form

x = wteλt

In this way, it may be natural to attempt to find a secondindependent solution

of the form

x = wteλt

but, doing this and substituting x in the system we find thatw = 0. Thus, we propose

x = wteλt + ueλt

and substituting this new x in the system we find the system

(A− λI) w = 0

(A− λI) u = w

The first equation is already solved with w = v and only thesecond one is remaining to be solved.

doing this and substituting x in the system we find thatw = 0. Thus, we propose

x = wteλt + ueλt

(A− λI) w = 0

(A− λI) u = w

but, doing this and

substituting x in the system we find thatw = 0. Thus, we propose

x = wteλt + ueλt

(A− λI) w = 0

(A− λI) u = w

but, doing this and substituting x in the system

we find thatw = 0. Thus, we propose

x = wteλt + ueλt

(A− λI) w = 0

(A− λI) u = w

but, doing this and substituting x in the system we find that

w = 0. Thus, we propose

x = wteλt + ueλt

(A− λI) w = 0

(A− λI) u = w

but, doing this and substituting x in the system we find thatw = 0. Thus,

we propose

x = wteλt + ueλt

(A− λI) w = 0

(A− λI) u = w

x = wteλt + ueλt

(A− λI) w = 0

(A− λI) u = w

x = wteλt + ueλt

(A− λI) w = 0

(A− λI) u = w

x = wteλt + ueλt

substituting this new x in the system we find the system

(A− λI) w = 0

(A− λI) u = w

x = wteλt + ueλt

and substituting

this new x in the system we find the system

(A− λI) w = 0

(A− λI) u = w

x = wteλt + ueλt

and substituting this new x

in the system we find the system

(A− λI) w = 0

(A− λI) u = w

x = wteλt + ueλt

and substituting this new x in the system

we find the system

(A− λI) w = 0

(A− λI) u = w

x = wteλt + ueλt

and substituting this new x in the system we find

the system

(A− λI) w = 0

(A− λI) u = w

x = wteλt + ueλt

(A− λI) w = 0

(A− λI) u = w

x = wteλt + ueλt

(A− λI) w = 0

(A− λI) u = w

x = wteλt + ueλt

(A− λI) w = 0

(A− λI) u = w

x = wteλt + ueλt

(A− λI) w = 0

(A− λI) u = w

The first equation

is already solved with w = v and only thesecond one is remaining to be solved.

x = wteλt + ueλt

(A− λI) w = 0

(A− λI) u = w

The first equation is already solved

with w = v and only thesecond one is remaining to be solved.

x = wteλt + ueλt

(A− λI) w = 0

(A− λI) u = w

The first equation is already solved with w = v and

only thesecond one is remaining to be solved.

x = wteλt + ueλt

(A− λI) w = 0

(A− λI) u = w

The first equation is already solved with w = v and only thesecond one

is remaining to be solved.

x = wteλt + ueλt

(A− λI) w = 0

(A− λI) u = w

The first equation is already solved with w = v and only thesecond one is remaining

to be solved.

x = wteλt + ueλt

(A− λI) w = 0

(A− λI) u = w

Example 9.8

Find the solution of the system

x′ = Ax =

(1 −11 3

Solution

|A− λI| =

∣∣∣∣1− λ −11 3− λ

∣∣∣∣ = 0

(λ− 1)(λ− 3) + 1 = 0 =⇒ (λ− 2)2 = 0

Example 9.8

x′ = Ax =

(1 −11 3

Solution

|A− λI| =

∣∣∣∣1− λ −11 3− λ

∣∣∣∣ = 0

(λ− 1)(λ− 3) + 1 = 0 =⇒ (λ− 2)2 = 0

Example 9.8

the solution of the system

x′ = Ax =

(1 −11 3

Solution

|A− λI| =

∣∣∣∣1− λ −11 3− λ

∣∣∣∣ = 0

(λ− 1)(λ− 3) + 1 = 0 =⇒ (λ− 2)2 = 0

Example 9.8

Find the solution

of the system

x′ = Ax =

(1 −11 3

Solution

|A− λI| =

∣∣∣∣1− λ −11 3− λ

∣∣∣∣ = 0

(λ− 1)(λ− 3) + 1 = 0 =⇒ (λ− 2)2 = 0

Example 9.8

x′ = Ax =

(1 −11 3

Solution

|A− λI| =

∣∣∣∣1− λ −11 3− λ

∣∣∣∣ = 0

(λ− 1)(λ− 3) + 1 = 0 =⇒ (λ− 2)2 = 0

Example 9.8

x′ = Ax =

(1 −11 3

Solution

|A− λI| =

∣∣∣∣1− λ −11 3− λ

∣∣∣∣ = 0

(λ− 1)(λ− 3) + 1 = 0 =⇒ (λ− 2)2 = 0

Example 9.8

x′ = Ax =

(1 −11 3

Solution

|A− λI| =

∣∣∣∣1− λ −11 3− λ

∣∣∣∣ = 0

(λ− 1)(λ− 3) + 1 = 0 =⇒ (λ− 2)2 = 0

Example 9.8

x′ = Ax =

(1 −11 3

Solution

|A− λI| =

∣∣∣∣1− λ −11 3− λ

∣∣∣∣ = 0

(λ− 1)(λ− 3) + 1 = 0 =⇒ (λ− 2)2 = 0

Example 9.8

x′ = Ax =

(1 −11 3

Solution

Let’s find

|A− λI| =

∣∣∣∣1− λ −11 3− λ

∣∣∣∣ = 0

(λ− 1)(λ− 3) + 1 = 0 =⇒ (λ− 2)2 = 0

Example 9.8

x′ = Ax =

(1 −11 3

Solution

of the matrix A

|A− λI| =

∣∣∣∣1− λ −11 3− λ

∣∣∣∣ = 0

(λ− 1)(λ− 3) + 1 = 0 =⇒ (λ− 2)2 = 0

Example 9.8

x′ = Ax =

(1 −11 3

Solution

|A− λI| =

∣∣∣∣1− λ −11 3− λ

∣∣∣∣ = 0

(λ− 1)(λ− 3) + 1 = 0 =⇒ (λ− 2)2 = 0

Example 9.8

x′ = Ax =

(1 −11 3

Solution

|A− λI| =

∣∣∣∣1− λ −11 3− λ

∣∣∣∣ = 0

(λ− 1)(λ− 3) + 1 = 0 =⇒ (λ− 2)2 = 0

Example 9.8

x′ = Ax =

(1 −11 3

Solution

|A− λI| =

∣∣∣∣1− λ −11 3− λ

∣∣∣∣ = 0

(λ− 1)(λ− 3) + 1 = 0 =⇒ (λ− 2)2 = 0

Example 9.8

x′ = Ax =

(1 −11 3

Solution

|A− λI| =

∣∣∣∣1− λ −11 3− λ

∣∣∣∣ = 0

(λ− 1)(λ− 3) + 1 = 0 =⇒ (λ− 2)2 = 0

Example 9.8

x′ = Ax =

(1 −11 3

Solution

|A− λI| =

∣∣∣∣1− λ −11 3− λ

∣∣∣∣ = 0

(λ− 1)(λ− 3) + 1 = 0 =⇒ (λ− 2)2 = 0

λ1 = 2, λ2 = 2,

If λ1,2 = 2, then

(A− λ1,2I) v =

(1− λ −1

1 3− λ

)(v1v2

(−1 −11 1

)(v1v2

)and a corresponding eigenvector is

v(1) =

λ1 = 2,

λ2 = 2,

If λ1,2 = 2, then

(A− λ1,2I) v =

(1− λ −1

1 3− λ

)(v1v2

(−1 −11 1

)(v1v2

v(1) =

λ1 = 2, λ2 = 2,

If λ1,2 = 2, then

(A− λ1,2I) v =

(1− λ −1

1 3− λ

)(v1v2

(−1 −11 1

)(v1v2

v(1) =

λ1 = 2, λ2 = 2,

λ1,2 = 2, then

(A− λ1,2I) v =

(1− λ −1

1 3− λ

)(v1v2

(−1 −11 1

)(v1v2

v(1) =

λ1 = 2, λ2 = 2,

If λ1,2 = 2,

(A− λ1,2I) v =

(1− λ −1

1 3− λ

)(v1v2

(−1 −11 1

)(v1v2

v(1) =

λ1 = 2, λ2 = 2,

If λ1,2 = 2, then

(A− λ1,2I) v =

(1− λ −1

1 3− λ

)(v1v2

(−1 −11 1

)(v1v2

v(1) =

λ1 = 2, λ2 = 2,

If λ1,2 = 2, then

(A− λ1,2I) v =

(1− λ −1

1 3− λ

)(v1v2

(−1 −11 1

)(v1v2

v(1) =

λ1 = 2, λ2 = 2,

If λ1,2 = 2, then

(A− λ1,2I) v =

(1− λ −1

1 3− λ

)(v1v2

(−1 −11 1

)(v1v2

v(1) =

λ1 = 2, λ2 = 2,

If λ1,2 = 2, then

(A− λ1,2I) v =

(1− λ −1

1 3− λ

)(v1v2

(−1 −11 1

)(v1v2

v(1) =

λ1 = 2, λ2 = 2,

If λ1,2 = 2, then

(A− λ1,2I) v =

(1− λ −1

1 3− λ

)(v1v2

(−1 −11 1

)(v1v2

v(1) =

λ1 = 2, λ2 = 2,

If λ1,2 = 2, then

(A− λ1,2I) v =

(1− λ −1

1 3− λ

)(v1v2

(−1 −11 1

)(v1v2

v(1) =

λ1 = 2, λ2 = 2,

If λ1,2 = 2, then

(A− λ1,2I) v =

(1− λ −1

1 3− λ

)(v1v2

(−1 −11 1

)(v1v2

)and a corresponding

eigenvector is

v(1) =

λ1 = 2, λ2 = 2,

If λ1,2 = 2, then

(A− λ1,2I) v =

(1− λ −1

1 3− λ

)(v1v2

(−1 −11 1

)(v1v2

v(1) =

λ1 = 2, λ2 = 2,

If λ1,2 = 2, then

(A− λ1,2I) v =

(1− λ −1

1 3− λ

)(v1v2

(−1 −11 1

)(v1v2

v(1) =

and the solution is

x(1) =

Now, for the second solution we propose

x(2) = vte2t + ue2t

where u satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

the solution is

x(1) =

x(2) = vte2t + ue2t

where u satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

and the solution

x(1) =

x(2) = vte2t + ue2t

where u satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

and the solution is

x(1) =

x(2) = vte2t + ue2t

where u satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

and the solution is

x(1) =

x(2) = vte2t + ue2t

where u satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

and the solution is

x(1) =

for the second solution we propose

x(2) = vte2t + ue2t

where u satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

and the solution is

x(1) =

Now, for the second

solution we propose

x(2) = vte2t + ue2t

where u satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

and the solution is

x(1) =

Now, for the second solution

we propose

x(2) = vte2t + ue2t

where u satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

and the solution is

x(1) =

x(2) = vte2t + ue2t

where u satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

and the solution is

x(1) =

x(2) = vte2t +

where u satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

and the solution is

x(1) =

x(2) = vte2t + ue2t

where u satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

and the solution is

x(1) =

x(2) = vte2t + ue2t

u satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

and the solution is

x(1) =

x(2) = vte2t + ue2t

where u

satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

and the solution is

x(1) =

x(2) = vte2t + ue2t

where u satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

and the solution is

x(1) =

x(2) = vte2t + ue2t

where u satisfies

(A− λI) u =

(A− 2I) u = v

(−1 −11 1

)(u1u2

and the solution is

x(1) =

x(2) = vte2t + ue2t

where u satisfies

(A− λI) u = (A− 2I) u =

(−1 −11 1

)(u1u2

and the solution is

x(1) =

x(2) = vte2t + ue2t

where u satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

and the solution is

x(1) =

x(2) = vte2t + ue2t

where u satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

we have

−u1 − u2 = 1

so if u1 = k , where k is arbitrary, then u2 = −k − 1. If we write

−1− k

(0−1

(1−1

)then by substituting for w and u, we obtain

x(2) =

(1−1

)te2t +

(0−1

)e2t + ke2t

(1−1

we have

−u1 − u2 = 1

−1− k

(0−1

(1−1

x(2) =

(1−1

)te2t +

(0−1

)e2t + ke2t

(1−1

we have

−u1 − u2 = 1

−1− k

(0−1

(1−1

x(2) =

(1−1

)te2t +

(0−1

)e2t + ke2t

(1−1

we have

−u1 − u2 = 1

if u1 = k , where k is arbitrary, then u2 = −k − 1. If we write

−1− k

(0−1

(1−1

x(2) =

(1−1

)te2t +

(0−1

)e2t + ke2t

(1−1

we have

−u1 − u2 = 1

so if u1 = k ,

where k is arbitrary, then u2 = −k − 1. If we write

−1− k

(0−1

(1−1

x(2) =

(1−1

)te2t +

(0−1

)e2t + ke2t

(1−1

we have

−u1 − u2 = 1

so if u1 = k , where k

is arbitrary, then u2 = −k − 1. If we write

−1− k

(0−1

(1−1

x(2) =

(1−1

)te2t +

(0−1

)e2t + ke2t

(1−1

we have

−u1 − u2 = 1

so if u1 = k , where k is arbitrary,

then u2 = −k − 1. If we write

−1− k

(0−1

(1−1

x(2) =

(1−1

)te2t +

(0−1

)e2t + ke2t

(1−1

we have

−u1 − u2 = 1

so if u1 = k , where k is arbitrary, then u2 = −k − 1.

If we write

−1− k

(0−1

(1−1

x(2) =

(1−1

)te2t +

(0−1

)e2t + ke2t

(1−1

we have

−u1 − u2 = 1

−1− k

(0−1

(1−1

x(2) =

(1−1

)te2t +

(0−1

)e2t + ke2t

(1−1

we have

−u1 − u2 = 1

−1− k

(0−1

(1−1

x(2) =

(1−1

)te2t +

(0−1

)e2t + ke2t

(1−1

we have

−u1 − u2 = 1

−1− k

(0−1

(1−1

x(2) =

(1−1

)te2t +

(0−1

)e2t + ke2t

(1−1

we have

−u1 − u2 = 1

−1− k

(0−1

(1−1

x(2) =

(1−1

)te2t +

(0−1

)e2t + ke2t

(1−1

we have

−u1 − u2 = 1

−1− k

(0−1

(1−1

then by substituting for w and u, we obtain

x(2) =

(1−1

)te2t +

(0−1

)e2t + ke2t

(1−1

we have

−u1 − u2 = 1

−1− k

(0−1

(1−1

by substituting for w and u, we obtain

x(2) =

(1−1

)te2t +

(0−1

)e2t + ke2t

(1−1

we have

−u1 − u2 = 1

−1− k

(0−1

(1−1

)then by substituting

for w and u, we obtain

x(2) =

(1−1

)te2t +

(0−1

)e2t + ke2t

(1−1

we have

−u1 − u2 = 1

−1− k

(0−1

(1−1

)then by substituting for w and

u, we obtain

x(2) =

(1−1

)te2t +

(0−1

)e2t + ke2t

(1−1

we have

−u1 − u2 = 1

−1− k

(0−1

(1−1

)then by substituting for w and u,

we obtain

x(2) =

(1−1

)te2t +

(0−1

)e2t + ke2t

(1−1

we have

−u1 − u2 = 1

−1− k

(0−1

(1−1

x(2) =

(1−1

)te2t +

(0−1

)e2t + ke2t

(1−1

we have

−u1 − u2 = 1

−1− k

(0−1

(1−1

x(2) =

(1−1

)te2t +

(0−1

)e2t + ke2t

(1−1

we have

−u1 − u2 = 1

−1− k

(0−1

(1−1

x(2) =

(1−1

)te2t +

(0−1

)e2t + ke2t

(1−1

we have

−u1 − u2 = 1

−1− k

(0−1

(1−1

x(2) =

(1−1

)te2t +

(0−1

)e2t +

we have

−u1 − u2 = 1

−1− k

(0−1

(1−1

x(2) =

(1−1

)te2t +

(0−1

)e2t + ke2t

(1−1

The last term above is merely a multiple of the first solutionx (1)(t) and may be ignored, but the first two terms constitute anew solution:

x(2) =

(1−1

)te2t +

(0−1

An elementary calculation shows that W [x (1), x (2)](t) = − e4t 6= 0and therefore

{x (1), x (2)

}form a fundamental set of solutions of

the system. The general solution is

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

The last term above

is merely a multiple of the first solutionx (1)(t) and may be ignored, but the first two terms constitute anew solution:

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

The last term above is merely

a multiple of the first solutionx (1)(t) and may be ignored, but the first two terms constitute anew solution:

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

The last term above is merely a multiple

of the first solutionx (1)(t) and may be ignored, but the first two terms constitute anew solution:

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

The last term above is merely a multiple of the first solution

x (1)(t) and may be ignored, but the first two terms constitute anew solution:

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

The last term above is merely a multiple of the first solutionx (1)(t) and

may be ignored, but the first two terms constitute anew solution:

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

The last term above is merely a multiple of the first solutionx (1)(t) and may be ignored,

but the first two terms constitute anew solution:

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

The last term above is merely a multiple of the first solutionx (1)(t) and may be ignored, but

the first two terms constitute anew solution:

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

The last term above is merely a multiple of the first solutionx (1)(t) and may be ignored, but the first

two terms constitute anew solution:

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

The last term above is merely a multiple of the first solutionx (1)(t) and may be ignored, but the first two terms

constitute anew solution:

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

The last term above is merely a multiple of the first solutionx (1)(t) and may be ignored, but the first two terms constitute

anew solution:

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

x(2) =

(1−1

)te2t +

(0−1

An elementary calculation

shows that W [x (1), x (2)](t) = − e4t 6= 0and therefore

{x (1), x (2)

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

x(2) =

(1−1

)te2t +

(0−1

An elementary calculation shows that W [x (1), x (2)](t) =

− e4t 6= 0and therefore

{x (1), x (2)

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

x(2) =

(1−1

)te2t +

(0−1

An elementary calculation shows that W [x (1), x (2)](t) = − e4t 6= 0and

therefore{x (1), x (2)

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

form a fundamental set of solutions ofthe system. The general solution is

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

}form a fundamental set

of solutions ofthe system. The general solution is

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

}form a fundamental set of solutions

ofthe system. The general solution is

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

the system.

The general solution is

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

the system. The general solution

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

x = c1x(1) +

c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

Consider again the system

x′ = Ax

and suppose that r = λ is a double eigenvalue of A, but, there isonly one corresponding eigenvector v. Then one solution is

x(1)(t) = veλtwhere v satisfies

(A− λI) v = 0

and a second solution is given by

x(2)(t) = vteλt + ueλt

where u satisfies

(A− λI) u = v

Consider again

the system

x′ = Ax

(A− λI) v = 0

where u satisfies

(A− λI) u = v

x′ = Ax

(A− λI) v = 0

where u satisfies

(A− λI) u = v

x′ = Ax

(A− λI) v = 0

where u satisfies

(A− λI) u = v

x′ = Ax

suppose that r = λ is a double eigenvalue of A, but, there isonly one corresponding eigenvector v. Then one solution is

(A− λI) v = 0

where u satisfies

(A− λI) u = v

x′ = Ax

and suppose that

r = λ is a double eigenvalue of A, but, there isonly one corresponding eigenvector v. Then one solution is

(A− λI) v = 0

where u satisfies

(A− λI) u = v

x′ = Ax

and suppose that r = λ

is a double eigenvalue of A, but, there isonly one corresponding eigenvector v. Then one solution is

(A− λI) v = 0

where u satisfies

(A− λI) u = v

x′ = Ax

and suppose that r = λ is a double eigenvalue

of A, but, there isonly one corresponding eigenvector v. Then one solution is

(A− λI) v = 0

where u satisfies

(A− λI) u = v

x′ = Ax

and suppose that r = λ is a double eigenvalue of A,

but, there isonly one corresponding eigenvector v. Then one solution is

(A− λI) v = 0

where u satisfies

(A− λI) u = v

x′ = Ax

and suppose that r = λ is a double eigenvalue of A, but, there isonly one

corresponding eigenvector v. Then one solution is

(A− λI) v = 0

where u satisfies

(A− λI) u = v

x′ = Ax

and suppose that r = λ is a double eigenvalue of A, but, there isonly one corresponding eigenvector v.

Then one solution is

(A− λI) v = 0

where u satisfies

(A− λI) u = v

x′ = Ax

and suppose that r = λ is a double eigenvalue of A, but, there isonly one corresponding eigenvector v. Then

one solution is

(A− λI) v = 0

where u satisfies

(A− λI) u = v

x′ = Ax

(A− λI) v = 0

where u satisfies

(A− λI) u = v

x′ = Ax

x(1)(t) = veλt

where v satisfies

(A− λI) v = 0

where u satisfies

(A− λI) u = v

x′ = Ax

x(1)(t) = veλtwhere

v satisfies

(A− λI) v = 0

where u satisfies

(A− λI) u = v

x′ = Ax

x(1)(t) = veλtwhere v

satisfies

(A− λI) v = 0

where u satisfies

(A− λI) u = v

x′ = Ax

(A− λI) v = 0

where u satisfies

(A− λI) u = v

x′ = Ax

(A− λI) v = 0

where u satisfies

(A− λI) u = v

x′ = Ax

(A− λI) v = 0

a second solution is given by

where u satisfies

(A− λI) u = v

x′ = Ax

(A− λI) v = 0

and a second solution

is given by

where u satisfies

(A− λI) u = v

x′ = Ax

(A− λI) v = 0

where u satisfies

(A− λI) u = v

x′ = Ax

(A− λI) v = 0

where u satisfies

(A− λI) u = v

x′ = Ax

(A− λI) v = 0

u satisfies

(A− λI) u = v

x′ = Ax

(A− λI) v = 0

where u

satisfies

(A− λI) u = v

x′ = Ax

(A− λI) v = 0

where u satisfies

(A− λI) u = v

x′ = Ax

(A− λI) v = 0

where u satisfies

(A− λI) u = vDr. Marco A Roque Sol Linear Algebra. Session 9

Even though |A− λI| = 0, it can be shown that it is alwayspossible to solve it for u ( Actually, there are infinetly solutions ) .Now, Using the above equation, together with the equation for v,we get

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

The vector u is known as a generalized eigenvector.

Even though

|A− λI| = 0, it can be shown that it is alwayspossible to solve it for u ( Actually, there are infinetly solutions ) .Now, Using the above equation, together with the equation for v,we get

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

Even though |A− λI| = 0,

it can be shown that it is alwayspossible to solve it for u ( Actually, there are infinetly solutions ) .Now, Using the above equation, together with the equation for v,we get

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

Even though |A− λI| = 0, it can be shown that

it is alwayspossible to solve it for u ( Actually, there are infinetly solutions ) .Now, Using the above equation, together with the equation for v,we get

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

Even though |A− λI| = 0, it can be shown that it is alwayspossible

to solve it for u ( Actually, there are infinetly solutions ) .Now, Using the above equation, together with the equation for v,we get

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

Even though |A− λI| = 0, it can be shown that it is alwayspossible to solve it

for u ( Actually, there are infinetly solutions ) .Now, Using the above equation, together with the equation for v,we get

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

Even though |A− λI| = 0, it can be shown that it is alwayspossible to solve it for u

( Actually, there are infinetly solutions ) .Now, Using the above equation, together with the equation for v,we get

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

Even though |A− λI| = 0, it can be shown that it is alwayspossible to solve it for u ( Actually,

there are infinetly solutions ) .Now, Using the above equation, together with the equation for v,we get

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

Even though |A− λI| = 0, it can be shown that it is alwayspossible to solve it for u ( Actually, there are

infinetly solutions ) .Now, Using the above equation, together with the equation for v,we get

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

Even though |A− λI| = 0, it can be shown that it is alwayspossible to solve it for u ( Actually, there are infinetly solutions ) .

Now, Using the above equation, together with the equation for v,we get

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

Even though |A− λI| = 0, it can be shown that it is alwayspossible to solve it for u ( Actually, there are infinetly solutions ) .Now,

Using the above equation, together with the equation for v,we get

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

Even though |A− λI| = 0, it can be shown that it is alwayspossible to solve it for u ( Actually, there are infinetly solutions ) .Now, Using the above equation,

together with the equation for v,we get

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

Even though |A− λI| = 0, it can be shown that it is alwayspossible to solve it for u ( Actually, there are infinetly solutions ) .Now, Using the above equation, together with

the equation for v,we get

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

Even though |A− λI| = 0, it can be shown that it is alwayspossible to solve it for u ( Actually, there are infinetly solutions ) .Now, Using the above equation, together with the equation

for v,we get

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

Even though |A− λI| = 0, it can be shown that it is alwayspossible to solve it for u ( Actually, there are infinetly solutions ) .Now, Using the above equation, together with the equation for v,

we get

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

(A− λI)

[(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

(A− λI) [(A− λI) u = v]

(A− λI)2 u =

(A− λI) v

(A− λI)2 u = 0

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u =

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

The vector

u is known as a generalized eigenvector.

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

The vector u

is known as a generalized eigenvector.

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

The vector u is known as

a generalized eigenvector.

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

Diagonalization.

Diagonalizable Matrices.

The basic reason why a system of linear (algebraic or differential)equations presents some difficulty is that the equations are usuallycoupled.

Hence, the equations in the system must be solved simultaneously.On the contrary, if the system is uncoupled, then each equationcan be solved independently of all the others.

Transforming the coupled system into an equivalent uncoupledsystem ( in which each equation contains only one unknownvariable ) corresponds to transforming the coefficient matrix A intoa diagonal matrix. Eigenvectors are useful in accomplishing such atransformation.

Diagonalization.

The basic reason

why a system of linear (algebraic or differential)equations presents some difficulty is that the equations are usuallycoupled.

Diagonalization.

The basic reason why a system

of linear (algebraic or differential)equations presents some difficulty is that the equations are usuallycoupled.

Diagonalization.

The basic reason why a system of linear

(algebraic or differential)equations presents some difficulty is that the equations are usuallycoupled.

Diagonalization.

The basic reason why a system of linear (algebraic or differential)

equations presents some difficulty is that the equations are usuallycoupled.

Diagonalization.

The basic reason why a system of linear (algebraic or differential)equations

presents some difficulty is that the equations are usuallycoupled.

Diagonalization.

The basic reason why a system of linear (algebraic or differential)equations presents some difficulty

is that the equations are usuallycoupled.

Diagonalization.

The basic reason why a system of linear (algebraic or differential)equations presents some difficulty is that

the equations are usuallycoupled.

Diagonalization.

The basic reason why a system of linear (algebraic or differential)equations presents some difficulty is that the equations

are usuallycoupled.

Diagonalization.

The basic reason why a system of linear (algebraic or differential)equations presents some difficulty is that the equations are usually

coupled.

Diagonalization.

Hence,

the equations in the system must be solved simultaneously.On the contrary, if the system is uncoupled, then each equationcan be solved independently of all the others.

Diagonalization.

Hence, the equations

in the system must be solved simultaneously.On the contrary, if the system is uncoupled, then each equationcan be solved independently of all the others.

Diagonalization.

Hence, the equations in the system

must be solved simultaneously.On the contrary, if the system is uncoupled, then each equationcan be solved independently of all the others.

Diagonalization.

Hence, the equations in the system must be solved

simultaneously.On the contrary, if the system is uncoupled, then each equationcan be solved independently of all the others.

Diagonalization.

Hence, the equations in the system must be solved simultaneously.

On the contrary, if the system is uncoupled, then each equationcan be solved independently of all the others.

Diagonalization.

Hence, the equations in the system must be solved simultaneously.On the contrary,

if the system is uncoupled, then each equationcan be solved independently of all the others.

Diagonalization.

Hence, the equations in the system must be solved simultaneously.On the contrary, if the system

is uncoupled, then each equationcan be solved independently of all the others.

Diagonalization.

Hence, the equations in the system must be solved simultaneously.On the contrary, if the system is uncoupled,

then each equationcan be solved independently of all the others.

Diagonalization.

Hence, the equations in the system must be solved simultaneously.On the contrary, if the system is uncoupled, then each equation

can be solved independently of all the others.

Diagonalization.

Hence, the equations in the system must be solved simultaneously.On the contrary, if the system is uncoupled, then each equationcan be solved

independently of all the others.

Diagonalization.

Hence, the equations in the system must be solved simultaneously.On the contrary, if the system is uncoupled, then each equationcan be solved independently

of all the others.

Diagonalization.

Transforming

the coupled system into an equivalent uncoupledsystem ( in which each equation contains only one unknownvariable ) corresponds to transforming the coefficient matrix A intoa diagonal matrix. Eigenvectors are useful in accomplishing such atransformation.

Diagonalization.

Transforming the coupled system

into an equivalent uncoupledsystem ( in which each equation contains only one unknownvariable ) corresponds to transforming the coefficient matrix A intoa diagonal matrix. Eigenvectors are useful in accomplishing such atransformation.

Diagonalization.

Transforming the coupled system into an equivalent

uncoupledsystem ( in which each equation contains only one unknownvariable ) corresponds to transforming the coefficient matrix A intoa diagonal matrix. Eigenvectors are useful in accomplishing such atransformation.

Diagonalization.

Transforming the coupled system into an equivalent uncoupledsystem

( in which each equation contains only one unknownvariable ) corresponds to transforming the coefficient matrix A intoa diagonal matrix. Eigenvectors are useful in accomplishing such atransformation.

Diagonalization.

Transforming the coupled system into an equivalent uncoupledsystem ( in which

each equation contains only one unknownvariable ) corresponds to transforming the coefficient matrix A intoa diagonal matrix. Eigenvectors are useful in accomplishing such atransformation.

Diagonalization.

Transforming the coupled system into an equivalent uncoupledsystem ( in which each equation

contains only one unknownvariable ) corresponds to transforming the coefficient matrix A intoa diagonal matrix. Eigenvectors are useful in accomplishing such atransformation.

Diagonalization.

Transforming the coupled system into an equivalent uncoupledsystem ( in which each equation contains only

one unknownvariable ) corresponds to transforming the coefficient matrix A intoa diagonal matrix. Eigenvectors are useful in accomplishing such atransformation.

Diagonalization.

Transforming the coupled system into an equivalent uncoupledsystem ( in which each equation contains only one unknownvariable )

corresponds to transforming the coefficient matrix A intoa diagonal matrix. Eigenvectors are useful in accomplishing such atransformation.

Diagonalization.

Transforming the coupled system into an equivalent uncoupledsystem ( in which each equation contains only one unknownvariable ) corresponds

to transforming the coefficient matrix A intoa diagonal matrix. Eigenvectors are useful in accomplishing such atransformation.

Diagonalization.

Transforming the coupled system into an equivalent uncoupledsystem ( in which each equation contains only one unknownvariable ) corresponds to transforming

the coefficient matrix A intoa diagonal matrix. Eigenvectors are useful in accomplishing such atransformation.

Diagonalization.

Transforming the coupled system into an equivalent uncoupledsystem ( in which each equation contains only one unknownvariable ) corresponds to transforming the coefficient matrix

A intoa diagonal matrix. Eigenvectors are useful in accomplishing such atransformation.

Diagonalization.

Transforming the coupled system into an equivalent uncoupledsystem ( in which each equation contains only one unknownvariable ) corresponds to transforming the coefficient matrix A

intoa diagonal matrix. Eigenvectors are useful in accomplishing such atransformation.

Diagonalization.

Transforming the coupled system into an equivalent uncoupledsystem ( in which each equation contains only one unknownvariable ) corresponds to transforming the coefficient matrix A intoa

diagonal matrix. Eigenvectors are useful in accomplishing such atransformation.

Diagonalization.

Transforming the coupled system into an equivalent uncoupledsystem ( in which each equation contains only one unknownvariable ) corresponds to transforming the coefficient matrix A intoa diagonal matrix.

Eigenvectors are useful in accomplishing such atransformation.

Diagonalization.

Transforming the coupled system into an equivalent uncoupledsystem ( in which each equation contains only one unknownvariable ) corresponds to transforming the coefficient matrix A intoa diagonal matrix. Eigenvectors

are useful in accomplishing such atransformation.

Diagonalization.

Transforming the coupled system into an equivalent uncoupledsystem ( in which each equation contains only one unknownvariable ) corresponds to transforming the coefficient matrix A intoa diagonal matrix. Eigenvectors are useful

in accomplishing such atransformation.

Diagonalization.

Transforming the coupled system into an equivalent uncoupledsystem ( in which each equation contains only one unknownvariable ) corresponds to transforming the coefficient matrix A intoa diagonal matrix. Eigenvectors are useful in accomplishing

such atransformation.

Diagonalization.

Let’s assume that the matrix A has n eigenvectors x(1), x(2), ...,x(n) linearly indepedent, then

Ax(1) = λ1x(1); Ax(2) = λ2x(2); ...Ax(n) = λnx(n)

and considering the matrix

x(1)1 · · · x

......

x(1)n · · · x

Diagonalization.

Let’s assume

that the matrix A has n eigenvectors x(1), x(2), ...,x(n) linearly indepedent, then

x(1)1 · · · x

......

x(1)n · · · x

Diagonalization.

Let’s assume that the matrix

A has n eigenvectors x(1), x(2), ...,x(n) linearly indepedent, then

x(1)1 · · · x

......

x(1)n · · · x

Diagonalization.

Let’s assume that the matrix A

has n eigenvectors x(1), x(2), ...,x(n) linearly indepedent, then

x(1)1 · · · x

......

x(1)n · · · x

Diagonalization.

Let’s assume that the matrix A has n eigenvectors x(1), x(2), ...,x(n)

linearly indepedent, then

x(1)1 · · · x

......

x(1)n · · · x

Diagonalization.

x(1)1 · · · x

......

x(1)n · · · x

Diagonalization.

x(1)1 · · · x

......

x(1)n · · · x

Diagonalization.

considering the matrix

x(1)1 · · · x

......

x(1)n · · · x

Diagonalization.

x(1)1 · · · x

......

x(1)n · · · x

Diagonalization.

x(1)1 · · · x

......

x(1)n · · · x

Diagonalization.

we have

Ax(1) · · · Ax

......

... · · ·...

(1)1 · · · λnx

λ1x(1)2

λ1x(1)n λnx

Diagonalization.

we have

Ax(1) · · · Ax

......

... · · ·...

(1)1 · · · λnx

λ1x(1)2

λ1x(1)n λnx

Diagonalization.

we have

Ax(1) · · · Ax

......

... · · ·...

(1)1 · · · λnx

λ1x(1)2

λ1x(1)n λnx

Diagonalization.

we have

Ax(1) · · · Ax

......

... · · ·...

(1)1 · · · λnx

λ1x(1)2

λ1x(1)n λnx

Diagonalization.

we have

Ax(1) · · · Ax

......

... · · ·...

(1)1 · · · λnx

λ1x(1)2

λ1x(1)n λnx

Diagonalization.

we have

Ax(1) · · · Ax

......

... · · ·...

(1)1 · · · λnx

λ1x(1)2

λ1x(1)n λnx

Diagonalization.

where D is the diagonal matrix

λ2. . .

whose diagonal elements are the eigenvalues of A. From the lastequations we have that it follows that

U−1AU = D ⇐⇒ A = UDU−1

Thus, if the eigenvalues and eigenvectors of A are known, A canbe transformed into a diagonal matrix by the process shown in theabove equation.

Diagonalization.

where D

is the diagonal matrix

λ2. . .

U−1AU = D ⇐⇒ A = UDU−1

Diagonalization.

λ2. . .

U−1AU = D ⇐⇒ A = UDU−1

Diagonalization.

λ2. . .

U−1AU = D ⇐⇒ A = UDU−1

Diagonalization.

λ2. . .

U−1AU = D ⇐⇒ A = UDU−1

Diagonalization.

λ2. . .

diagonal elements are the eigenvalues of A. From the lastequations we have that it follows that

U−1AU = D ⇐⇒ A = UDU−1

Diagonalization.

λ2. . .

whose diagonal elements

are the eigenvalues of A. From the lastequations we have that it follows that

U−1AU = D ⇐⇒ A = UDU−1

Diagonalization.

λ2. . .

whose diagonal elements are the eigenvalues

of A. From the lastequations we have that it follows that

U−1AU = D ⇐⇒ A = UDU−1

Diagonalization.

λ2. . .

whose diagonal elements are the eigenvalues of A.

From the lastequations we have that it follows that

U−1AU = D ⇐⇒ A = UDU−1

Diagonalization.

λ2. . .

whose diagonal elements are the eigenvalues of A. From the lastequations

we have that it follows that

U−1AU = D ⇐⇒ A = UDU−1

Diagonalization.

λ2. . .

whose diagonal elements are the eigenvalues of A. From the lastequations we have that

it follows that

U−1AU = D ⇐⇒ A = UDU−1

Diagonalization.

λ2. . .

U−1AU = D ⇐⇒ A = UDU−1

Diagonalization.

λ2. . .

U−1AU = D ⇐⇒ A = UDU−1

Diagonalization.

λ2. . .

U−1AU = D ⇐⇒ A = UDU−1

if the eigenvalues and eigenvectors of A are known, A canbe transformed into a diagonal matrix by the process shown in theabove equation.

Diagonalization.

λ2. . .

U−1AU = D ⇐⇒ A = UDU−1

Thus, if the eigenvalues and

eigenvectors of A are known, A canbe transformed into a diagonal matrix by the process shown in theabove equation.

Diagonalization.

λ2. . .

U−1AU = D ⇐⇒ A = UDU−1

Thus, if the eigenvalues and eigenvectors

of A are known, A canbe transformed into a diagonal matrix by the process shown in theabove equation.

Diagonalization.

λ2. . .

U−1AU = D ⇐⇒ A = UDU−1

Thus, if the eigenvalues and eigenvectors of A

are known, A canbe transformed into a diagonal matrix by the process shown in theabove equation.

Diagonalization.

λ2. . .

U−1AU = D ⇐⇒ A = UDU−1

Thus, if the eigenvalues and eigenvectors of A are known,

A canbe transformed into a diagonal matrix by the process shown in theabove equation.

Diagonalization.

λ2. . .

U−1AU = D ⇐⇒ A = UDU−1

Thus, if the eigenvalues and eigenvectors of A are known, A canbe transformed

into a diagonal matrix by the process shown in theabove equation.

Diagonalization.

λ2. . .

U−1AU = D ⇐⇒ A = UDU−1

Thus, if the eigenvalues and eigenvectors of A are known, A canbe transformed into a

diagonal matrix by the process shown in theabove equation.

Diagonalization.

λ2. . .

U−1AU = D ⇐⇒ A = UDU−1

Thus, if the eigenvalues and eigenvectors of A are known, A canbe transformed into a diagonal matrix

by the process shown in theabove equation.

Diagonalization.

λ2. . .

U−1AU = D ⇐⇒ A = UDU−1

Thus, if the eigenvalues and eigenvectors of A are known, A canbe transformed into a diagonal matrix by the process

shown in theabove equation.

Diagonalization.

λ2. . .

U−1AU = D ⇐⇒ A = UDU−1

Thus, if the eigenvalues and eigenvectors of A are known, A canbe transformed into a diagonal matrix by the process shown in

theabove equation.

Diagonalization.

λ2. . .

U−1AU = D ⇐⇒ A = UDU−1

Diagonalization.

This process is known as a similarity transformation.Alternatively, we may say that A is diagonalizable.

If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).

Finally, we note that if A has fewer than n linearly independenteigenvectors, then there is no matrix U such that U−1AU = D. Inthis case, A is not diagonalizable.

Diagonalization.

This process

is known as a similarity transformation.Alternatively, we may say that A is diagonalizable.

Diagonalization.

This process is known as

a similarity transformation.Alternatively, we may say that A is diagonalizable.

Diagonalization.

This process is known as a similarity transformation.

Alternatively, we may say that A is diagonalizable.

Diagonalization.

This process is known as a similarity transformation.Alternatively,

we may say that A is diagonalizable.

Diagonalization.

This process is known as a similarity transformation.Alternatively, we may say

that A is diagonalizable.

Diagonalization.

This process is known as a similarity transformation.Alternatively, we may say that A

is diagonalizable.

Diagonalization.

A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).

Diagonalization.

is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).

Diagonalization.

If A is Hermitian

( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).

Diagonalization.

If A is Hermitian ( A = (A∗)T ),

then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).

Diagonalization.

If A is Hermitian ( A = (A∗)T ), then the determination

of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).

Diagonalization.

If A is Hermitian ( A = (A∗)T ), then the determination of U−1

isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).

Diagonalization.

If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple.

The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).

Diagonalization.

If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n)

of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).

Diagonalization.

If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A

are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).

Diagonalization.

If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known

to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).

Diagonalization.

If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to be

mutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).

Diagonalization.

If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal,

so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).

Diagonalization.

If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us

choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).

Diagonalization.

If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them

so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).

Diagonalization.

If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that

they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).

Diagonalization.

If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are

alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).

Diagonalization.

If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized

by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).

Diagonalization.

If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1

for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).

Diagonalization.

If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i .

It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).

Diagonalization.

If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy

verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).

Diagonalization.

If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify that

U−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).

Diagonalization.

If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗.

In other words, the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).

Diagonalization.

If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words,

the inverse of U is the same as itsadjoint (the transpose of its complex conjugate).

Diagonalization.

If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse

of U is the same as itsadjoint (the transpose of its complex conjugate).

Diagonalization.

If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U

is the same as itsadjoint (the transpose of its complex conjugate).

Diagonalization.

If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same

as itsadjoint (the transpose of its complex conjugate).

Diagonalization.

If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint

(the transpose of its complex conjugate).

Diagonalization.

If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose

of its complex conjugate).

Diagonalization.

If A is Hermitian ( A = (A∗)T ), then the determination of U−1 isvery simple. The eigenvectors v(1), ..., v(n) of A are known to bemutually orthogonal, so let us choose them so that they are alsonormalized by < v(i), v(i) >= 1 for each i . It is easy verify thatU−1 = U∗. In other words, the inverse of U is the same as itsadjoint (the transpose of its

complex conjugate).

Diagonalization.

Finally,

we note that if A has fewer than n linearly independenteigenvectors, then there is no matrix U such that U−1AU = D. Inthis case, A is not diagonalizable.

Diagonalization.

Finally, we note

that if A has fewer than n linearly independenteigenvectors, then there is no matrix U such that U−1AU = D. Inthis case, A is not diagonalizable.

Diagonalization.

Finally, we note that if A

has fewer than n linearly independenteigenvectors, then there is no matrix U such that U−1AU = D. Inthis case, A is not diagonalizable.

Diagonalization.

Finally, we note that if A has fewer

than n linearly independenteigenvectors, then there is no matrix U such that U−1AU = D. Inthis case, A is not diagonalizable.

Diagonalization.

Finally, we note that if A has fewer than n

linearly independenteigenvectors, then there is no matrix U such that U−1AU = D. Inthis case, A is not diagonalizable.

Diagonalization.

Finally, we note that if A has fewer than n linearly independenteigenvectors,

then there is no matrix U such that U−1AU = D. Inthis case, A is not diagonalizable.

Diagonalization.

Finally, we note that if A has fewer than n linearly independenteigenvectors, then there is no

matrix U such that U−1AU = D. Inthis case, A is not diagonalizable.

Diagonalization.

Finally, we note that if A has fewer than n linearly independenteigenvectors, then there is no matrix U

such that U−1AU = D. Inthis case, A is not diagonalizable.

Diagonalization.

Finally, we note that if A has fewer than n linearly independenteigenvectors, then there is no matrix U such that

U−1AU = D. Inthis case, A is not diagonalizable.

Diagonalization.

Finally, we note that if A has fewer than n linearly independenteigenvectors, then there is no matrix U such that U−1AU = D.

Inthis case, A is not diagonalizable.

Diagonalization.

Finally, we note that if A has fewer than n linearly independenteigenvectors, then there is no matrix U such that U−1AU = D. Inthis case,

A is not diagonalizable.

Diagonalization.

Finally, we note that if A has fewer than n linearly independenteigenvectors, then there is no matrix U such that U−1AU = D. Inthis case, A

is not diagonalizable.

Diagonalization.

Finally, we note that if A has fewer than n linearly independenteigenvectors, then there is no matrix U such that U−1AU = D. Inthis case, A is not

diagonalizable.

Diagonalization.

Fundamental Matrices

Let’s start with the system

x′ = P(t)x

Suppose that x(1)(t), ..., x(n)(t) form a fundamental set ofsolutions on some interval α < t < β. Then the matrix

Ψ(t) =

x(1)1 · · · x

......

x(1)n · · · x

whose columns are the vectors x(1)(t), ..., x(n)(t), is said to be afundamental matrix for the linear system. Since the set ofsolutions is linearly independent the matrix is nonsingular.

Diagonalization.

x′ = P(t)x

Ψ(t) =

x(1)1 · · · x

......

x(1)n · · · x

Diagonalization.

x′ = P(t)x

Ψ(t) =

x(1)1 · · · x

......

x(1)n · · · x

Diagonalization.

x′ = P(t)x

Ψ(t) =

x(1)1 · · · x

......

x(1)n · · · x

Diagonalization.

x′ = P(t)x

Suppose that x(1)(t), ..., x(n)(t)

form a fundamental set ofsolutions on some interval α < t < β. Then the matrix

Ψ(t) =

x(1)1 · · · x

......

x(1)n · · · x

Diagonalization.

x′ = P(t)x

Suppose that x(1)(t), ..., x(n)(t) form a fundamental set

ofsolutions on some interval α < t < β. Then the matrix

Ψ(t) =

x(1)1 · · · x

......

x(1)n · · · x

Diagonalization.

x′ = P(t)x

Suppose that x(1)(t), ..., x(n)(t) form a fundamental set ofsolutions

on some interval α < t < β. Then the matrix

Ψ(t) =

x(1)1 · · · x

......

x(1)n · · · x

Diagonalization.

x′ = P(t)x

Suppose that x(1)(t), ..., x(n)(t) form a fundamental set ofsolutions on some interval

α < t < β. Then the matrix

Ψ(t) =

x(1)1 · · · x

......

x(1)n · · · x

Diagonalization.

x′ = P(t)x

Suppose that x(1)(t), ..., x(n)(t) form a fundamental set ofsolutions on some interval α < t < β.

Then the matrix

Ψ(t) =

x(1)1 · · · x

......

x(1)n · · · x

Diagonalization.

x′ = P(t)x

Ψ(t) =

x(1)1 · · · x

......

x(1)n · · · x

Diagonalization.

x′ = P(t)x

Ψ(t) =

x(1)1 · · · x

......

x(1)n · · · x

Diagonalization.

x′ = P(t)x

Ψ(t) =

x(1)1 · · · x

......

x(1)n · · · x

Diagonalization.

x′ = P(t)x

Ψ(t) =

x(1)1 · · · x

......

x(1)n · · · x

whose columns

are the vectors x(1)(t), ..., x(n)(t), is said to be afundamental matrix for the linear system. Since the set ofsolutions is linearly independent the matrix is nonsingular.

Diagonalization.

x′ = P(t)x

Ψ(t) =

x(1)1 · · · x

......

x(1)n · · · x

whose columns are the vectors

x(1)(t), ..., x(n)(t), is said to be afundamental matrix for the linear system. Since the set ofsolutions is linearly independent the matrix is nonsingular.

Diagonalization.

x′ = P(t)x

Ψ(t) =

x(1)1 · · · x

......

x(1)n · · · x

whose columns are the vectors x(1)(t), ..., x(n)(t),

is said to be afundamental matrix for the linear system. Since the set ofsolutions is linearly independent the matrix is nonsingular.

Diagonalization.

x′ = P(t)x

Ψ(t) =

x(1)1 · · · x

......

x(1)n · · · x

whose columns are the vectors x(1)(t), ..., x(n)(t), is said to be

afundamental matrix for the linear system. Since the set ofsolutions is linearly independent the matrix is nonsingular.

Diagonalization.

x′ = P(t)x

Ψ(t) =

x(1)1 · · · x

......

x(1)n · · · x

whose columns are the vectors x(1)(t), ..., x(n)(t), is said to be afundamental matrix

for the linear system. Since the set ofsolutions is linearly independent the matrix is nonsingular.

Diagonalization.

x′ = P(t)x

Ψ(t) =

x(1)1 · · · x

......

x(1)n · · · x

whose columns are the vectors x(1)(t), ..., x(n)(t), is said to be afundamental matrix for the linear system.

Since the set ofsolutions is linearly independent the matrix is nonsingular.

Diagonalization.

x′ = P(t)x

Ψ(t) =

x(1)1 · · · x

......

x(1)n · · · x

whose columns are the vectors x(1)(t), ..., x(n)(t), is said to be afundamental matrix for the linear system. Since the set ofsolutions

is linearly independent the matrix is nonsingular.

Diagonalization.

x′ = P(t)x

Ψ(t) =

x(1)1 · · · x

......

x(1)n · · · x

whose columns are the vectors x(1)(t), ..., x(n)(t), is said to be afundamental matrix for the linear system. Since the set ofsolutions is linearly independent

the matrix is nonsingular.

Diagonalization.

x′ = P(t)x

Ψ(t) =

x(1)1 · · · x

......

x(1)n · · · x

whose columns are the vectors x(1)(t), ..., x(n)(t), is said to be afundamental matrix for the linear system. Since the set ofsolutions is linearly independent the matrix

is nonsingular.

Diagonalization.

x′ = P(t)x

Ψ(t) =

x(1)1 · · · x

......

x(1)n · · · x

Diagonalization.

Thus, for example, a fundamental matrix for the system

x′ = Ax =

(1 −11 3

can be formed from the solutions x (1)(t) and x (2)(t):

x(1) =

)e2t ; x(2) =

(1−1

)te2t +

(0−1

Diagonalization.

for example, a fundamental matrix for the system

x′ = Ax =

(1 −11 3

x(1) =

)e2t ; x(2) =

(1−1

)te2t +

(0−1

Diagonalization.

Thus, for example,

a fundamental matrix for the system

x′ = Ax =

(1 −11 3

x(1) =

)e2t ; x(2) =

(1−1

)te2t +

(0−1

Diagonalization.

Thus, for example, a fundamental matrix

for the system

x′ = Ax =

(1 −11 3

x(1) =

)e2t ; x(2) =

(1−1

)te2t +

(0−1

Diagonalization.

x′ = Ax =

(1 −11 3

x(1) =

)e2t ; x(2) =

(1−1

)te2t +

(0−1

Diagonalization.

x′ = Ax =

(1 −11 3

x(1) =

)e2t ; x(2) =

(1−1

)te2t +

(0−1

Diagonalization.

x′ = Ax =

(1 −11 3

x(1) =

)e2t ; x(2) =

(1−1

)te2t +

(0−1

Diagonalization.

x′ = Ax =

(1 −11 3

can be formed

from the solutions x (1)(t) and x (2)(t):

x(1) =

)e2t ; x(2) =

(1−1

)te2t +

(0−1

Diagonalization.

x′ = Ax =

(1 −11 3

can be formed from the solutions

x (1)(t) and x (2)(t):

x(1) =

)e2t ; x(2) =

(1−1

)te2t +

(0−1

Diagonalization.

x′ = Ax =

(1 −11 3

can be formed from the solutions x (1)(t) and

x (2)(t):

x(1) =

)e2t ; x(2) =

(1−1

)te2t +

(0−1

Diagonalization.

x′ = Ax =

(1 −11 3

x(1) =

)e2t ; x(2) =

(1−1

)te2t +

(0−1

Diagonalization.

x′ = Ax =

(1 −11 3

x(1) =

)e2t ;

x(2) =

(1−1

)te2t +

(0−1

Diagonalization.

x′ = Ax =

(1 −11 3

x(1) =

)e2t ; x(2) =

(1−1

)te2t +

(0−1

Diagonalization.

x′ = Ax =

(1 −11 3

x(1) =

)e2t ; x(2) =

(1−1

)te2t +

(0−1

Diagonalization.

x′ = Ax =

(1 −11 3

x(1) =

)e2t ; x(2) =

(1−1

)te2t +

(0−1

Diagonalization.

x′ = Ax =

(1 −11 3

x(1) =

)e2t ; x(2) =

(1−1

)te2t +

(0−1

Diagonalization.

Ψ(t) =

(e2t te2t

−e2t −te2t − e2t

)= e2t

(1 t−1 −1− t

)Recall that each column of the fundamental matrix Ψ(t) is asolution of the ODE. It follows that Ψ(t) satisfies the matrixdifferential equation

Ψ′ = P(t)Ψ

In particular, the fundamental matrix Φ(t) that satisfies Φ(0) = Iis called the special fundamental matrix and can also be foundfrom the relation Φ(t) = Ψ(t)Ψ−1(0).

Diagonalization.

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

Ψ′ = P(t)Ψ

Diagonalization.

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

Ψ′ = P(t)Ψ

Diagonalization.

Ψ(t) =

(e2t te2t

1 t−1 −1− t

Ψ′ = P(t)Ψ

Diagonalization.

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

Ψ′ = P(t)Ψ

Diagonalization.

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

Recall that each column of the fundamental matrix Ψ(t) is asolution of the ODE. It follows that Ψ(t) satisfies the matrixdifferential equation

Ψ′ = P(t)Ψ

Diagonalization.

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

)Recall that each column

of the fundamental matrix Ψ(t) is asolution of the ODE. It follows that Ψ(t) satisfies the matrixdifferential equation

Ψ′ = P(t)Ψ

Diagonalization.

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

)Recall that each column of the fundamental matrix Ψ(t)

is asolution of the ODE. It follows that Ψ(t) satisfies the matrixdifferential equation

Ψ′ = P(t)Ψ

Diagonalization.

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

)Recall that each column of the fundamental matrix Ψ(t) is asolution of the ODE.

It follows that Ψ(t) satisfies the matrixdifferential equation

Ψ′ = P(t)Ψ

Diagonalization.

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

)Recall that each column of the fundamental matrix Ψ(t) is asolution of the ODE. It follows that Ψ(t)

satisfies the matrixdifferential equation

Ψ′ = P(t)Ψ

Diagonalization.

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

Ψ′ = P(t)Ψ

Diagonalization.

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

Ψ′ = P(t)Ψ

Diagonalization.

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

Ψ′ = P(t)Ψ

In particular,

the fundamental matrix Φ(t) that satisfies Φ(0) = Iis called the special fundamental matrix and can also be foundfrom the relation Φ(t) = Ψ(t)Ψ−1(0).

Diagonalization.

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

Ψ′ = P(t)Ψ

In particular, the fundamental matrix Φ(t) that satisfies Φ(0) = I

is called the special fundamental matrix and can also be foundfrom the relation Φ(t) = Ψ(t)Ψ−1(0).

Diagonalization.

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

Ψ′ = P(t)Ψ

In particular, the fundamental matrix Φ(t) that satisfies Φ(0) = Iis called the special fundamental matrix and

can also be foundfrom the relation Φ(t) = Ψ(t)Ψ−1(0).

Diagonalization.

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

Ψ′ = P(t)Ψ

In particular, the fundamental matrix Φ(t) that satisfies Φ(0) = Iis called the special fundamental matrix and can also be found

from the relation Φ(t) = Ψ(t)Ψ−1(0).

Diagonalization.

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

Ψ′ = P(t)Ψ

In particular, the fundamental matrix Φ(t) that satisfies Φ(0) = Iis called the special fundamental matrix and can also be foundfrom the relation

Φ(t) = Ψ(t)Ψ−1(0).

Diagonalization.

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

Ψ′ = P(t)Ψ

In particular, the fundamental matrix Φ(t) that satisfies Φ(0) = Iis called the special fundamental matrix and can also be foundfrom the relation Φ(t) =

Ψ(t)Ψ−1(0).

Diagonalization.

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

Ψ′ = P(t)Ψ

In particular, the fundamental matrix Φ(t) that satisfies Φ(0) = Iis called the special fundamental matrix and can also be foundfrom the relation Φ(t) = Ψ(t)

Ψ−1(0).

Diagonalization.

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

Ψ′ = P(t)Ψ

Diagonalization.

Thus, in this case

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

Φ(t) = e2t(

1− t −tt 1 + t

)The latter matrix is also known as the exponential matrix eAt .

Diagonalization.

Thus, in this case

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

Φ(t) = e2t(

1− t −tt 1 + t

Diagonalization.

Thus, in this case

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

Φ(t) = e2t(

1− t −tt 1 + t

Diagonalization.

Thus, in this case

Ψ(0) =

(1 0−1 −1

=⇒ Ψ−1(0) =

(1 0−1 −1

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

Φ(t) = e2t(

1− t −tt 1 + t

Diagonalization.

Thus, in this case

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

Φ(t) = e2t(

1− t −tt 1 + t

Diagonalization.

Thus, in this case

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

Φ(t) = e2t(

1− t −tt 1 + t

Diagonalization.

Thus, in this case

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

Φ(t) =

Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

Φ(t) = e2t(

1− t −tt 1 + t

Diagonalization.

Thus, in this case

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

Φ(t) = Ψ(t)Ψ−1(0) =

1 t−1 −1− t

) (1 0−1 −1

Φ(t) = e2t(

1− t −tt 1 + t

Diagonalization.

Thus, in this case

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

(1 0−1 −1

Φ(t) = e2t(

1− t −tt 1 + t

Diagonalization.

Thus, in this case

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

Φ(t) = e2t(

1− t −tt 1 + t

Diagonalization.

Thus, in this case

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

Φ(t) =

1− t −tt 1 + t

Diagonalization.

Thus, in this case

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

Φ(t) = e2t(

1− t −tt 1 + t

The latter matrix is also known as the exponential matrix eAt .

Diagonalization.

Thus, in this case

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

Φ(t) = e2t(

1− t −tt 1 + t

)The latter matrix

is also known as the exponential matrix eAt .

Diagonalization.

Thus, in this case

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

Φ(t) = e2t(

1− t −tt 1 + t

)The latter matrix is also known

as the exponential matrix eAt .

Diagonalization.

Thus, in this case

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

Φ(t) = e2t(

1− t −tt 1 + t

)The latter matrix is also known as the

exponential matrix eAt .

Diagonalization.

Thus, in this case

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

Φ(t) = e2t(

1− t −tt 1 + t

Diagonalization.

The Matrix eAt

Recall that the solution of the initial value problem

x ′ = ax , x(0) = x0, a = constant

is given by

x(t) = x0eat

Now, consider the corresponding initial value problem for an n × nsystem

x′ = Ax, x(0) = x0

where A is a constant matrix.

Diagonalization.

The Matrix eAt

is given by

x(t) = x0eat

x′ = Ax, x(0) = x0

Diagonalization.

The Matrix eAt

is given by

x(t) = x0eat

x′ = Ax, x(0) = x0

Diagonalization.

The Matrix eAt

is given by

x(t) = x0eat

x′ = Ax, x(0) = x0

Diagonalization.

The Matrix eAt

is given by

x(t) = x0eat

x′ = Ax, x(0) = x0

Diagonalization.

The Matrix eAt

is given by

x(t) = x0eat

x′ = Ax, x(0) = x0

Diagonalization.

The Matrix eAt

is given by

x(t) = x0eat

Now, consider the corresponding initial value problem

for an n × nsystem

x′ = Ax, x(0) = x0

Diagonalization.

The Matrix eAt

is given by

x(t) = x0eat

x′ = Ax, x(0) = x0

Diagonalization.

The Matrix eAt

is given by

x(t) = x0eat

x′ = Ax, x(0) = x0

Diagonalization.

The Matrix eAt

is given by

x(t) = x0eat

x′ = Ax, x(0) = x0

Diagonalization.

Applying the results of already obtained, we can write its solutionas

x = Φ(t)x0

where Φ(0) = I. Thus, Φ(t), is playing the roll of eat . let’s seethis with more detail.

The scalar exponential function eat can be represented by thepower series

eat = 1 +∞∑n=1

which converges for all t. Let us now replace the scalar a by then × n constant matrix A and consider the corresponding series

Diagonalization.

Applying the results of already obtained,

we can write its solutionas

x = Φ(t)x0

eat = 1 +∞∑n=1

Diagonalization.

x = Φ(t)x0

eat = 1 +∞∑n=1

Diagonalization.

x = Φ(t)x0

eat = 1 +∞∑n=1

Diagonalization.

x = Φ(t)x0

where Φ(0) = I.

Thus, Φ(t), is playing the roll of eat . let’s seethis with more detail.

eat = 1 +∞∑n=1

Diagonalization.

x = Φ(t)x0

where Φ(0) = I. Thus, Φ(t), is playing the roll of eat .

let’s seethis with more detail.

eat = 1 +∞∑n=1

Diagonalization.

x = Φ(t)x0

eat = 1 +∞∑n=1

Diagonalization.

x = Φ(t)x0

The scalar exponential function eat

can be represented by thepower series

eat = 1 +∞∑n=1

Diagonalization.

x = Φ(t)x0

eat = 1 +∞∑n=1

Diagonalization.

x = Φ(t)x0

eat = 1 +∞∑n=1

Diagonalization.

x = Φ(t)x0

eat = 1 +∞∑n=1

which converges for all t.

Let us now replace the scalar a by then × n constant matrix A and consider the corresponding series

Diagonalization.

x = Φ(t)x0

eat = 1 +∞∑n=1

which converges for all t. Let us now replace the scalar a by then × n constant matrix A and

consider the corresponding series

Diagonalization.

x = Φ(t)x0

eat = 1 +∞∑n=1

Diagonalization.

I +∞∑n=1

n!= I + At +

2!+ ...+

n!+ ...

Each term in the series is an n × n matrix. It is possible to showthat each element of this matrix sum converges for all t asn→∞. Thus, we have a well defined n × n matrix, which will bedenote by eAt

eAt = I +∞∑n=1

By differentiating the above series term by term, we obtain

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

Diagonalization.

I +∞∑n=1

I + At +A2t2

2!+ ...+

n!+ ...

eAt = I +∞∑n=1

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

Diagonalization.

I +∞∑n=1

n!= I +

At +A2t2

2!+ ...+

n!+ ...

eAt = I +∞∑n=1

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

Diagonalization.

I +∞∑n=1

n!= I + At +

2!+ ...+

n!+ ...

eAt = I +∞∑n=1

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

Diagonalization.

I +∞∑n=1

n!= I + At +

...+Ant2

n!+ ...

eAt = I +∞∑n=1

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

Diagonalization.

I +∞∑n=1

n!= I + At +

2!+ ...+

n!+ ...

eAt = I +∞∑n=1

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

Diagonalization.

I +∞∑n=1

n!= I + At +

2!+ ...+

eAt = I +∞∑n=1

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

Diagonalization.

I +∞∑n=1

n!= I + At +

2!+ ...+

n!+ ...

eAt = I +∞∑n=1

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

Diagonalization.

I +∞∑n=1

n!= I + At +

2!+ ...+

n!+ ...

Each term in the series is an n × n matrix.

It is possible to showthat each element of this matrix sum converges for all t asn→∞. Thus, we have a well defined n × n matrix, which will bedenote by eAt

eAt = I +∞∑n=1

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

Diagonalization.

I +∞∑n=1

n!= I + At +

2!+ ...+

n!+ ...

Each term in the series is an n × n matrix. It is possible to showthat

each element of this matrix sum converges for all t asn→∞. Thus, we have a well defined n × n matrix, which will bedenote by eAt

eAt = I +∞∑n=1

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

Diagonalization.

I +∞∑n=1

n!= I + At +

2!+ ...+

n!+ ...

Each term in the series is an n × n matrix. It is possible to showthat each element of this matrix sum converges

for all t asn→∞. Thus, we have a well defined n × n matrix, which will bedenote by eAt

eAt = I +∞∑n=1

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

Diagonalization.

I +∞∑n=1

n!= I + At +

2!+ ...+

n!+ ...

Each term in the series is an n × n matrix. It is possible to showthat each element of this matrix sum converges for all t asn→∞.

Thus, we have a well defined n × n matrix, which will bedenote by eAt

eAt = I +∞∑n=1

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

Diagonalization.

I +∞∑n=1

n!= I + At +

2!+ ...+

n!+ ...

Each term in the series is an n × n matrix. It is possible to showthat each element of this matrix sum converges for all t asn→∞. Thus, we have a well defined n × n matrix,

which will bedenote by eAt

eAt = I +∞∑n=1

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

Diagonalization.

I +∞∑n=1

n!= I + At +

2!+ ...+

n!+ ...

eAt = I +∞∑n=1

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

Diagonalization.

I +∞∑n=1

n!= I + At +

2!+ ...+

n!+ ...

eAt = I +∞∑n=1

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

Diagonalization.

I +∞∑n=1

n!= I + At +

2!+ ...+

n!+ ...

eAt = I +∞∑n=1

By differentiating the above series

term by term, we obtain

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

Diagonalization.

I +∞∑n=1

n!= I + At +

2!+ ...+

n!+ ...

eAt = I +∞∑n=1

By differentiating the above series term by term,

we obtain

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

Diagonalization.

I +∞∑n=1

n!= I + At +

2!+ ...+

n!+ ...

eAt = I +∞∑n=1

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

Diagonalization.

I +∞∑n=1

n!= I + At +

2!+ ...+

n!+ ...

eAt = I +∞∑n=1

∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

Diagonalization.

I +∞∑n=1

n!= I + At +

2!+ ...+

n!+ ...

eAt = I +∞∑n=1

=∞∑n=1

Antn−1

(n − 1)!=

∞∑n=1

]= AeAt

Diagonalization.

I +∞∑n=1

n!= I + At +

2!+ ...+

n!+ ...

eAt = I +∞∑n=1

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

Diagonalization.

I +∞∑n=1

n!= I + At +

2!+ ...+

n!+ ...

eAt = I +∞∑n=1

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

Diagonalization.

Therefore, eAt satisfies the differential equation

= AeAt

Further, by setting t = 0 in the definition of eAt we find that eAt

satisfies the initial condition

eAt∣∣∣t=0

In this way, we have that the special fundamental matrix Φsatisfies the same initial value problem as eAt , namely,

Φ′ = AΦ, Φ(0) = I

Diagonalization.

= AeAt

eAt∣∣∣t=0

Φ′ = AΦ, Φ(0) = I

Diagonalization.

= AeAt

eAt∣∣∣t=0

Φ′ = AΦ, Φ(0) = I

Diagonalization.

= AeAt

Further, by setting t = 0

in the definition of eAt we find that eAt

eAt∣∣∣t=0

Φ′ = AΦ, Φ(0) = I

Diagonalization.

= AeAt

Further, by setting t = 0 in the definition of eAt

we find that eAt

eAt∣∣∣t=0

Φ′ = AΦ, Φ(0) = I

Diagonalization.

= AeAt

eAt∣∣∣t=0

Φ′ = AΦ, Φ(0) = I

Diagonalization.

= AeAt

eAt∣∣∣t=0

Φ′ = AΦ, Φ(0) = I

Diagonalization.

= AeAt

eAt∣∣∣t=0

In this way,

we have that the special fundamental matrix Φsatisfies the same initial value problem as eAt , namely,

Φ′ = AΦ, Φ(0) = I

Diagonalization.

= AeAt

eAt∣∣∣t=0

In this way, we have that

the special fundamental matrix Φsatisfies the same initial value problem as eAt , namely,

Φ′ = AΦ, Φ(0) = I

Diagonalization.

= AeAt

eAt∣∣∣t=0

In this way, we have that the special fundamental matrix Φ

satisfies the same initial value problem as eAt , namely,

Φ′ = AΦ, Φ(0) = I

Diagonalization.

= AeAt

eAt∣∣∣t=0

In this way, we have that the special fundamental matrix Φsatisfies the same initial value problem as

eAt , namely,

Φ′ = AΦ, Φ(0) = I

Diagonalization.

= AeAt

eAt∣∣∣t=0

Φ′ = AΦ, Φ(0) = I

Diagonalization.

= AeAt

eAt∣∣∣t=0

Φ′ = AΦ, Φ(0) = I

Diagonalization.

Then, by uniqueness of an IVP (extended to matrix differentialequations), we conclude that eAt and the special fundamentalmatrix Φ(t) are the same. Thus, we can write the solution of theinitial value problem

x = Ax, x(0) = x0

in the form

x = eAtx0

Diagonalization.

Then, by uniqueness of an IVP

(extended to matrix differentialequations), we conclude that eAt and the special fundamentalmatrix Φ(t) are the same. Thus, we can write the solution of theinitial value problem

x = Ax, x(0) = x0

in the form

x = eAtx0

Diagonalization.

Then, by uniqueness of an IVP (extended to matrix differentialequations),

we conclude that eAt and the special fundamentalmatrix Φ(t) are the same. Thus, we can write the solution of theinitial value problem

x = Ax, x(0) = x0

in the form

x = eAtx0

Diagonalization.

Then, by uniqueness of an IVP (extended to matrix differentialequations), we conclude that eAt and

the special fundamentalmatrix Φ(t) are the same. Thus, we can write the solution of theinitial value problem

x = Ax, x(0) = x0

in the form

x = eAtx0

Diagonalization.

Then, by uniqueness of an IVP (extended to matrix differentialequations), we conclude that eAt and the special fundamentalmatrix

Φ(t) are the same. Thus, we can write the solution of theinitial value problem

x = Ax, x(0) = x0

in the form

x = eAtx0

Diagonalization.

Then, by uniqueness of an IVP (extended to matrix differentialequations), we conclude that eAt and the special fundamentalmatrix Φ(t)

are the same. Thus, we can write the solution of theinitial value problem

x = Ax, x(0) = x0

in the form

x = eAtx0

Diagonalization.

Then, by uniqueness of an IVP (extended to matrix differentialequations), we conclude that eAt and the special fundamentalmatrix Φ(t) are the same.

Thus, we can write the solution of theinitial value problem

x = Ax, x(0) = x0

in the form

x = eAtx0

Diagonalization.

Then, by uniqueness of an IVP (extended to matrix differentialequations), we conclude that eAt and the special fundamentalmatrix Φ(t) are the same. Thus,

we can write the solution of theinitial value problem

x = Ax, x(0) = x0

in the form

x = eAtx0

Diagonalization.

Then, by uniqueness of an IVP (extended to matrix differentialequations), we conclude that eAt and the special fundamentalmatrix Φ(t) are the same. Thus, we can write

the solution of theinitial value problem

x = Ax, x(0) = x0

in the form

x = eAtx0

Diagonalization.

Then, by uniqueness of an IVP (extended to matrix differentialequations), we conclude that eAt and the special fundamentalmatrix Φ(t) are the same. Thus, we can write the solution

of theinitial value problem

x = Ax, x(0) = x0

in the form

x = eAtx0

Diagonalization.

Then, by uniqueness of an IVP (extended to matrix differentialequations), we conclude that eAt and the special fundamentalmatrix Φ(t) are the same. Thus, we can write the solution of theinitial

value problem

x = Ax, x(0) = x0

in the form

x = eAtx0

Diagonalization.

x = Ax, x(0) = x0

in the form

x = eAtx0

Diagonalization.

x = Ax, x(0) = x0

in the form

x = eAtx0

Diagonalization.

x = Ax, x(0) = x0

in the form

x = eAtx0

Diagonalization.

x = Ax, x(0) = x0

in the form

x = eAtx0

Diagonalization.

Jordan Canonical Forms

An n × n matrix A can be diagonalized only if it has a fullcomplement of n linearly independent eigenvectors.

If there is a shortage of eigenvectors (because of repeatedeigenvalues), then A can always be transformed into a nearlydiagonal matrix called its Jordan form.

Diagonalization.

n × n matrix A can be diagonalized only if it has a fullcomplement of n linearly independent eigenvectors.

Diagonalization.

An n × n matrix

A can be diagonalized only if it has a fullcomplement of n linearly independent eigenvectors.

Diagonalization.

An n × n matrix A

can be diagonalized only if it has a fullcomplement of n linearly independent eigenvectors.

Diagonalization.

An n × n matrix A can be diagonalized

only if it has a fullcomplement of n linearly independent eigenvectors.

Diagonalization.

An n × n matrix A can be diagonalized only if

it has a fullcomplement of n linearly independent eigenvectors.

Diagonalization.

An n × n matrix A can be diagonalized only if it has

a fullcomplement of n linearly independent eigenvectors.

Diagonalization.

An n × n matrix A can be diagonalized only if it has a fullcomplement

of n linearly independent eigenvectors.

Diagonalization.

An n × n matrix A can be diagonalized only if it has a fullcomplement of

n linearly independent eigenvectors.

Diagonalization.

An n × n matrix A can be diagonalized only if it has a fullcomplement of n

linearly independent eigenvectors.

Diagonalization.

An n × n matrix A can be diagonalized only if it has a fullcomplement of n linearly independent

eigenvectors.

Diagonalization.

If there is a shortage of eigenvectors

(because of repeatedeigenvalues), then A can always be transformed into a nearlydiagonal matrix called its Jordan form.

Diagonalization.

If there is a shortage of eigenvectors (because of repeatedeigenvalues),

then A can always be transformed into a nearlydiagonal matrix called its Jordan form.

Diagonalization.

If there is a shortage of eigenvectors (because of repeatedeigenvalues), then A can always be transformed

into a nearlydiagonal matrix called its Jordan form.

Diagonalization.

If there is a shortage of eigenvectors (because of repeatedeigenvalues), then A can always be transformed into a nearlydiagonal matrix called

its Jordan form.

Diagonalization.

A Jordan form, J, has the eigenvalues of A on the main diagonal,ones in certain positions on the diagonal above the main diagonal,and zeros elsewhere.

J(t) =

λ1 10 λ1 10 0 λ1

λ2 10 λ2

λ3. . .

λn 10 λn

Diagonalization.

A Jordan form,

J, has the eigenvalues of A on the main diagonal,ones in certain positions on the diagonal above the main diagonal,and zeros elsewhere.

J(t) =

λ1 10 λ1 10 0 λ1

λ2 10 λ2

λ3. . .

λn 10 λn

Diagonalization.

A Jordan form, J,

has the eigenvalues of A on the main diagonal,ones in certain positions on the diagonal above the main diagonal,and zeros elsewhere.

J(t) =

λ1 10 λ1 10 0 λ1

λ2 10 λ2

λ3. . .

λn 10 λn

Diagonalization.

A Jordan form, J, has the eigenvalues

of A on the main diagonal,ones in certain positions on the diagonal above the main diagonal,and zeros elsewhere.

J(t) =

λ1 10 λ1 10 0 λ1

λ2 10 λ2

λ3. . .

λn 10 λn

Diagonalization.

A Jordan form, J, has the eigenvalues of A

on the main diagonal,ones in certain positions on the diagonal above the main diagonal,and zeros elsewhere.

J(t) =

λ1 10 λ1 10 0 λ1

λ2 10 λ2

λ3. . .

λn 10 λn

Diagonalization.

A Jordan form, J, has the eigenvalues of A on the

main diagonal,ones in certain positions on the diagonal above the main diagonal,and zeros elsewhere.

J(t) =

λ1 10 λ1 10 0 λ1

λ2 10 λ2

λ3. . .

λn 10 λn

Diagonalization.

A Jordan form, J, has the eigenvalues of A on the main diagonal,

ones in certain positions on the diagonal above the main diagonal,and zeros elsewhere.

J(t) =

λ1 10 λ1 10 0 λ1

λ2 10 λ2

λ3. . .

λn 10 λn

Diagonalization.

A Jordan form, J, has the eigenvalues of A on the main diagonal,ones in certain positions

on the diagonal above the main diagonal,and zeros elsewhere.

J(t) =

λ1 10 λ1 10 0 λ1

λ2 10 λ2

λ3. . .

λn 10 λn

Diagonalization.

A Jordan form, J, has the eigenvalues of A on the main diagonal,ones in certain positions on the diagonal

above the main diagonal,and zeros elsewhere.

J(t) =

λ1 10 λ1 10 0 λ1

λ2 10 λ2

λ3. . .

λn 10 λn

Diagonalization.

A Jordan form, J, has the eigenvalues of A on the main diagonal,ones in certain positions on the diagonal above the main diagonal,and

zeros elsewhere.

J(t) =

λ1 10 λ1 10 0 λ1

λ2 10 λ2

λ3. . .

λn 10 λn

Diagonalization.

J(t) =

λ1 10 λ1 10 0 λ1

λ2 10 λ2

λ3. . .

λn 10 λn

Diagonalization.

J(t) =

λ1 10 λ1 10 0 λ1

λ2 10 λ2

λ3. . .

λn 10 λn

Diagonalization.

J(t) =

λ1 10 λ1 10 0 λ1

λ2 10 λ2

λ3. . .

λn 10 λn

Diagonalization.

J(t) =

λ1 10 λ1 10 0 λ1

λ2 10 λ2

λ3. . .

λn 10 λn

Diagonalization.

J(t) =

λ1 10 λ1 10 0 λ1

λ2 10 λ2

λn 10 λn

Diagonalization.

J(t) =

λ1 10 λ1 10 0 λ1

λ2 10 λ2

λ3. . .

λn 10 λn

Diagonalization.

J(t) =

λ1 10 λ1 10 0 λ1

λ2 10 λ2

λ3. . .

λn 10 λn

Diagonalization.

Consider again the matrix A given by the equation

x′ = Ax =

(1 −11 3

To transform A into its Jordan form, we construct thetransformation matrix U with the single eigenvector v in its firstcolumn and the generalized eigenvector u ( k = 0 ) in thesecond column. Then U and its inverse are given by

(1 0−1 −1

)U−1 =

(1 0−1 −1

)It follows that

J = U−1AU =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

Diagonalization.

Consider again

the matrix A given by the equation

x′ = Ax =

(1 −11 3

(1 0−1 −1

)U−1 =

(1 0−1 −1

)It follows that

J = U−1AU =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

Diagonalization.

Consider again the matrix A

given by the equation

x′ = Ax =

(1 −11 3

(1 0−1 −1

)U−1 =

(1 0−1 −1

)It follows that

J = U−1AU =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

Diagonalization.

x′ = Ax =

(1 −11 3

(1 0−1 −1

)U−1 =

(1 0−1 −1

)It follows that

J = U−1AU =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

Diagonalization.

x′ = Ax =

(1 −11 3

(1 0−1 −1

)U−1 =

(1 0−1 −1

)It follows that

J = U−1AU =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

Diagonalization.

x′ = Ax =

(1 −11 3

(1 0−1 −1

)U−1 =

(1 0−1 −1

)It follows that

J = U−1AU =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

Diagonalization.

x′ = Ax =

(1 −11 3

To transform

A into its Jordan form, we construct thetransformation matrix U with the single eigenvector v in its firstcolumn and the generalized eigenvector u ( k = 0 ) in thesecond column. Then U and its inverse are given by

(1 0−1 −1

)U−1 =

(1 0−1 −1

)It follows that

J = U−1AU =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

Diagonalization.

x′ = Ax =

(1 −11 3

To transform A into

its Jordan form, we construct thetransformation matrix U with the single eigenvector v in its firstcolumn and the generalized eigenvector u ( k = 0 ) in thesecond column. Then U and its inverse are given by

(1 0−1 −1

)U−1 =

(1 0−1 −1

)It follows that

J = U−1AU =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

Diagonalization.

x′ = Ax =

(1 −11 3

To transform A into its Jordan form,

we construct thetransformation matrix U with the single eigenvector v in its firstcolumn and the generalized eigenvector u ( k = 0 ) in thesecond column. Then U and its inverse are given by

(1 0−1 −1

)U−1 =

(1 0−1 −1

)It follows that

J = U−1AU =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

Diagonalization.

x′ = Ax =

(1 −11 3

To transform A into its Jordan form, we construct

thetransformation matrix U with the single eigenvector v in its firstcolumn and the generalized eigenvector u ( k = 0 ) in thesecond column. Then U and its inverse are given by

(1 0−1 −1

)U−1 =

(1 0−1 −1

)It follows that

J = U−1AU =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

Diagonalization.

x′ = Ax =

(1 −11 3

To transform A into its Jordan form, we construct thetransformation matrix

U with the single eigenvector v in its firstcolumn and the generalized eigenvector u ( k = 0 ) in thesecond column. Then U and its inverse are given by

(1 0−1 −1

)U−1 =

(1 0−1 −1

)It follows that

J = U−1AU =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

Diagonalization.

x′ = Ax =

(1 −11 3

To transform A into its Jordan form, we construct thetransformation matrix U

with the single eigenvector v in its firstcolumn and the generalized eigenvector u ( k = 0 ) in thesecond column. Then U and its inverse are given by

(1 0−1 −1

)U−1 =

(1 0−1 −1

)It follows that

J = U−1AU =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

Diagonalization.

x′ = Ax =

(1 −11 3

To transform A into its Jordan form, we construct thetransformation matrix U with the single eigenvector

v in its firstcolumn and the generalized eigenvector u ( k = 0 ) in thesecond column. Then U and its inverse are given by

(1 0−1 −1

)U−1 =

(1 0−1 −1

)It follows that

J = U−1AU =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

Diagonalization.

x′ = Ax =

(1 −11 3

To transform A into its Jordan form, we construct thetransformation matrix U with the single eigenvector v

in its firstcolumn and the generalized eigenvector u ( k = 0 ) in thesecond column. Then U and its inverse are given by

(1 0−1 −1

)U−1 =

(1 0−1 −1

)It follows that

J = U−1AU =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

Diagonalization.

x′ = Ax =

(1 −11 3

To transform A into its Jordan form, we construct thetransformation matrix U with the single eigenvector v in its firstcolumn and

the generalized eigenvector u ( k = 0 ) in thesecond column. Then U and its inverse are given by

(1 0−1 −1

)U−1 =

(1 0−1 −1

)It follows that

J = U−1AU =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

Diagonalization.

x′ = Ax =

(1 −11 3

To transform A into its Jordan form, we construct thetransformation matrix U with the single eigenvector v in its firstcolumn and the generalized eigenvector

u ( k = 0 ) in thesecond column. Then U and its inverse are given by

(1 0−1 −1

)U−1 =

(1 0−1 −1

)It follows that

J = U−1AU =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

Diagonalization.

x′ = Ax =

(1 −11 3

To transform A into its Jordan form, we construct thetransformation matrix U with the single eigenvector v in its firstcolumn and the generalized eigenvector u ( k = 0 )

in thesecond column. Then U and its inverse are given by

(1 0−1 −1

)U−1 =

(1 0−1 −1

)It follows that

J = U−1AU =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

Diagonalization.

x′ = Ax =

(1 −11 3

To transform A into its Jordan form, we construct thetransformation matrix U with the single eigenvector v in its firstcolumn and the generalized eigenvector u ( k = 0 ) in thesecond column.

Then U and its inverse are given by

(1 0−1 −1

)U−1 =

(1 0−1 −1

)It follows that

J = U−1AU =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

Diagonalization.

x′ = Ax =

(1 −11 3

To transform A into its Jordan form, we construct thetransformation matrix U with the single eigenvector v in its firstcolumn and the generalized eigenvector u ( k = 0 ) in thesecond column. Then

U and its inverse are given by

(1 0−1 −1

)U−1 =

(1 0−1 −1

)It follows that

J = U−1AU =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

Diagonalization.

x′ = Ax =

(1 −11 3

To transform A into its Jordan form, we construct thetransformation matrix U with the single eigenvector v in its firstcolumn and the generalized eigenvector u ( k = 0 ) in thesecond column. Then U and

its inverse are given by

(1 0−1 −1

)U−1 =

(1 0−1 −1

)It follows that

J = U−1AU =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

Diagonalization.

x′ = Ax =

(1 −11 3

To transform A into its Jordan form, we construct thetransformation matrix U with the single eigenvector v in its firstcolumn and the generalized eigenvector u ( k = 0 ) in thesecond column. Then U and its inverse

are given by

(1 0−1 −1

)U−1 =

(1 0−1 −1

)It follows that

J = U−1AU =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

Diagonalization.

x′ = Ax =

(1 −11 3

(1 0−1 −1

)U−1 =

(1 0−1 −1

)It follows that

J = U−1AU =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

Diagonalization.

x′ = Ax =

(1 −11 3

(1 0−1 −1

)U−1 =

(1 0−1 −1

)It follows that

J = U−1AU =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

Diagonalization.

x′ = Ax =

(1 −11 3

(1 0−1 −1

U−1 =

(1 0−1 −1

)It follows that

J = U−1AU =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

Diagonalization.

x′ = Ax =

(1 −11 3

(1 0−1 −1

)U−1 =

(1 0−1 −1

)It follows that

J = U−1AU =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

Diagonalization.

x′ = Ax =

(1 −11 3

(1 0−1 −1

)U−1 =

(1 0−1 −1

It follows that

J = U−1AU =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

Diagonalization.

x′ = Ax =

(1 −11 3

(1 0−1 −1

)U−1 =

(1 0−1 −1

)It follows that

J = U−1AU =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

Diagonalization.

x′ = Ax =

(1 −11 3

(1 0−1 −1

)U−1 =

(1 0−1 −1

)It follows that

U−1AU =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

Diagonalization.

x′ = Ax =

(1 −11 3

(1 0−1 −1

)U−1 =

(1 0−1 −1

)It follows that

J = U−1AU =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

Diagonalization.

x′ = Ax =

(1 −11 3

(1 0−1 −1

)U−1 =

(1 0−1 −1

)It follows that

J = U−1AU =

(1 0−1 −1

(1 −11 3

) (1 0−1 −1

(2 10 2

Diagonalization.

x′ = Ax =

(1 −11 3

(1 0−1 −1

)U−1 =

(1 0−1 −1

)It follows that

J = U−1AU =

(1 0−1 −1

) (1 −11 3

(1 0−1 −1

(2 10 2

Diagonalization.

x′ = Ax =

(1 −11 3

(1 0−1 −1

)U−1 =

(1 0−1 −1

)It follows that

J = U−1AU =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

Diagonalization.

x′ = Ax =

(1 −11 3

(1 0−1 −1

)U−1 =

(1 0−1 −1

)It follows that

J = U−1AU =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

)Dr. Marco A Roque Sol Linear Algebra. Session 9

Diagonalization.

Finally, If we start again from

x′ = Ax =

(1 −11 3

the transformation x = Uy where U is given above, produces thesystem

y′ = Jy

y ′1 = 2y1 + y2, y ′2 = 2y2

y2 = c1e2t , y1 = c1te

2t + c2e2t

Diagonalization.

Finally,

If we start again from

x′ = Ax =

(1 −11 3

y′ = Jy

y ′1 = 2y1 + y2, y ′2 = 2y2

y2 = c1e2t , y1 = c1te

2t + c2e2t

Diagonalization.

x′ = Ax =

(1 −11 3

y′ = Jy

y ′1 = 2y1 + y2, y ′2 = 2y2

y2 = c1e2t , y1 = c1te

2t + c2e2t

Diagonalization.

x′ = Ax =

(1 −11 3

y′ = Jy

y ′1 = 2y1 + y2, y ′2 = 2y2

y2 = c1e2t , y1 = c1te

2t + c2e2t

Diagonalization.

x′ = Ax =

(1 −11 3

y′ = Jy

y ′1 = 2y1 + y2, y ′2 = 2y2

y2 = c1e2t , y1 = c1te

2t + c2e2t

Diagonalization.

x′ = Ax =

(1 −11 3

the transformation

x = Uy where U is given above, produces thesystem

y′ = Jy

y ′1 = 2y1 + y2, y ′2 = 2y2

y2 = c1e2t , y1 = c1te

2t + c2e2t

Diagonalization.

x′ = Ax =

(1 −11 3

the transformation x = Uy

where U is given above, produces thesystem

y′ = Jy

y ′1 = 2y1 + y2, y ′2 = 2y2

y2 = c1e2t , y1 = c1te

2t + c2e2t

Diagonalization.

x′ = Ax =

(1 −11 3

the transformation x = Uy where U

is given above, produces thesystem

y′ = Jy

y ′1 = 2y1 + y2, y ′2 = 2y2

y2 = c1e2t , y1 = c1te

2t + c2e2t

Diagonalization.

x′ = Ax =

(1 −11 3

the transformation x = Uy where U is given above,

produces thesystem

y′ = Jy

y ′1 = 2y1 + y2, y ′2 = 2y2

y2 = c1e2t , y1 = c1te

2t + c2e2t

Diagonalization.

x′ = Ax =

(1 −11 3

y′ = Jy

y ′1 = 2y1 + y2, y ′2 = 2y2

y2 = c1e2t , y1 = c1te

2t + c2e2t

Diagonalization.

x′ = Ax =

(1 −11 3

y′ = Jy

y ′1 = 2y1 + y2, y ′2 = 2y2

y2 = c1e2t , y1 = c1te

2t + c2e2t

Diagonalization.

x′ = Ax =

(1 −11 3

y′ = Jy

y ′1 = 2y1 + y2,

y ′2 = 2y2

y2 = c1e2t , y1 = c1te

2t + c2e2t

Diagonalization.

x′ = Ax =

(1 −11 3

y′ = Jy

y ′1 = 2y1 + y2, y ′2 = 2y2

y2 = c1e2t , y1 = c1te

2t + c2e2t

Diagonalization.

x′ = Ax =

(1 −11 3

y′ = Jy

y ′1 = 2y1 + y2, y ′2 = 2y2

y2 = c1e2t ,

y1 = c1te2t + c2e

Diagonalization.

x′ = Ax =

(1 −11 3

y′ = Jy

y ′1 = 2y1 + y2, y ′2 = 2y2

y2 = c1e2t , y1 = c1te

2t + c2e2t

Diagonalization.

Thus, two independent solutions of the y−system are

y(1)(t) =

)e2t ; y(2)(t) =

and the corresponding fundamental matrix is

Ψ̂(t) =

(e2t te2t

)Since Ψ̂(0) = I, we can also identify this matrix as eJt . To obtaina fundamental matrix for the original system, we now form theproduct

Ψ(t) = UeJt =

(e2t te2t

−e2t −e2t − te2t

)which is the same as the fundamental matrix that we obtainedbefore.

Diagonalization.

two independent solutions of the y−system are

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

Ψ(t) = UeJt =

(e2t te2t

Diagonalization.

Thus, two independent solutions

of the y−system are

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

Ψ(t) = UeJt =

(e2t te2t

Diagonalization.

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

Ψ(t) = UeJt =

(e2t te2t

Diagonalization.

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

Ψ(t) = UeJt =

(e2t te2t

Diagonalization.

y(1)(t) =

)e2t ;

y(2)(t) =

Ψ̂(t) =

(e2t te2t

Ψ(t) = UeJt =

(e2t te2t

Diagonalization.

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

Ψ(t) = UeJt =

(e2t te2t

Diagonalization.

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

Ψ(t) = UeJt =

(e2t te2t

Diagonalization.

y(1)(t) =

)e2t ; y(2)(t) =

and the corresponding

fundamental matrix is

Ψ̂(t) =

(e2t te2t

Ψ(t) = UeJt =

(e2t te2t

Diagonalization.

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

Ψ(t) = UeJt =

(e2t te2t

Diagonalization.

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

Ψ(t) = UeJt =

(e2t te2t

Diagonalization.

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

Since Ψ̂(0) = I, we can also identify this matrix as eJt . To obtaina fundamental matrix for the original system, we now form theproduct

Ψ(t) = UeJt =

(e2t te2t

Diagonalization.

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

)Since Ψ̂(0) = I,

we can also identify this matrix as eJt . To obtaina fundamental matrix for the original system, we now form theproduct

Ψ(t) = UeJt =

(e2t te2t

Diagonalization.

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

)Since Ψ̂(0) = I, we can also

identify this matrix as eJt . To obtaina fundamental matrix for the original system, we now form theproduct

Ψ(t) = UeJt =

(e2t te2t

Diagonalization.

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

)Since Ψ̂(0) = I, we can also identify this matrix as eJt .

To obtaina fundamental matrix for the original system, we now form theproduct

Ψ(t) = UeJt =

(e2t te2t

Diagonalization.

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

)Since Ψ̂(0) = I, we can also identify this matrix as eJt . To obtain

a fundamental matrix for the original system, we now form theproduct

Ψ(t) = UeJt =

(e2t te2t

Diagonalization.

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

)Since Ψ̂(0) = I, we can also identify this matrix as eJt . To obtaina fundamental matrix

for the original system, we now form theproduct

Ψ(t) = UeJt =

(e2t te2t

Diagonalization.

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

)Since Ψ̂(0) = I, we can also identify this matrix as eJt . To obtaina fundamental matrix for the original system,

we now form theproduct

Ψ(t) = UeJt =

(e2t te2t

Diagonalization.

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

)Since Ψ̂(0) = I, we can also identify this matrix as eJt . To obtaina fundamental matrix for the original system, we now form

theproduct

Ψ(t) = UeJt =

(e2t te2t

Diagonalization.

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

Ψ(t) = UeJt =

(e2t te2t

Diagonalization.

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

Ψ(t) =

UeJt =

(e2t te2t

Diagonalization.

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

Ψ(t) = UeJt =

(e2t te2t

Diagonalization.

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

Ψ(t) = UeJt =

(e2t te2t

which is the same as the fundamental matrix that we obtainedbefore.

Diagonalization.

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

Ψ(t) = UeJt =

(e2t te2t

)which is t

he same as the fundamental matrix that we obtainedbefore.

Diagonalization.

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

Ψ(t) = UeJt =

(e2t te2t

)which is the same

as the fundamental matrix that we obtainedbefore.

Diagonalization.

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

Ψ(t) = UeJt =

(e2t te2t

)which is the same as the

fundamental matrix that we obtainedbefore.

Diagonalization.

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

Ψ(t) = UeJt =

(e2t te2t

)which is the same as the fundamental matrix

that we obtainedbefore.

Diagonalization.

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

Ψ(t) = UeJt =

(e2t te2t

SVD. IntroductionSVD. Examples

SVD. Introduction

Singular Value DecompositionIn this section, we assume throughout that A is an m × n matrixwith m ≥ n. (This assumption is made for convenience only; allthe results will also hold if m < n).

We will present a method for determining how close A is to amatrix of smaller rank. The method involves factoring A into aproduct UΣV T , where U is an m ×m orthogonal matrix, V is ann × n orthogonal matrix, and Σ is an m × n matrix whoseoff-diagonal entries are all 0′s and whose diagonal elements satisfy

σ1 ≥ σ2 ≥ · · · ≥ σn ≥ 0

SVD. Introduction

Singular Value DecompositionIn this section,

we assume throughout that A is an m × n matrixwith m ≥ n. (This assumption is made for convenience only; allthe results will also hold if m < n).

σ1 ≥ σ2 ≥ · · · ≥ σn ≥ 0

SVD. Introduction

Singular Value DecompositionIn this section, we assume throughout that A

is an m × n matrixwith m ≥ n. (This assumption is made for convenience only; allthe results will also hold if m < n).

σ1 ≥ σ2 ≥ · · · ≥ σn ≥ 0

SVD. Introduction

Singular Value DecompositionIn this section, we assume throughout that A is an m × n matrix

with m ≥ n. (This assumption is made for convenience only; allthe results will also hold if m < n).

σ1 ≥ σ2 ≥ · · · ≥ σn ≥ 0

SVD. Introduction

Singular Value DecompositionIn this section, we assume throughout that A is an m × n matrixwith m ≥ n.

(This assumption is made for convenience only; allthe results will also hold if m < n).

σ1 ≥ σ2 ≥ · · · ≥ σn ≥ 0

SVD. Introduction

Singular Value DecompositionIn this section, we assume throughout that A is an m × n matrixwith m ≥ n. (This assumption

is made for convenience only; allthe results will also hold if m < n).

σ1 ≥ σ2 ≥ · · · ≥ σn ≥ 0

SVD. Introduction

Singular Value DecompositionIn this section, we assume throughout that A is an m × n matrixwith m ≥ n. (This assumption is made for

convenience only; allthe results will also hold if m < n).

σ1 ≥ σ2 ≥ · · · ≥ σn ≥ 0

SVD. Introduction

Singular Value DecompositionIn this section, we assume throughout that A is an m × n matrixwith m ≥ n. (This assumption is made for convenience only;

allthe results will also hold if m < n).

σ1 ≥ σ2 ≥ · · · ≥ σn ≥ 0

SVD. Introduction

Singular Value DecompositionIn this section, we assume throughout that A is an m × n matrixwith m ≥ n. (This assumption is made for convenience only; allthe results

will also hold if m < n).

σ1 ≥ σ2 ≥ · · · ≥ σn ≥ 0

SVD. Introduction

Singular Value DecompositionIn this section, we assume throughout that A is an m × n matrixwith m ≥ n. (This assumption is made for convenience only; allthe results will also hold

if m < n).

σ1 ≥ σ2 ≥ · · · ≥ σn ≥ 0

SVD. Introduction

σ1 ≥ σ2 ≥ · · · ≥ σn ≥ 0

SVD. Introduction

We will present

a method for determining how close A is to amatrix of smaller rank. The method involves factoring A into aproduct UΣV T , where U is an m ×m orthogonal matrix, V is ann × n orthogonal matrix, and Σ is an m × n matrix whoseoff-diagonal entries are all 0′s and whose diagonal elements satisfy

σ1 ≥ σ2 ≥ · · · ≥ σn ≥ 0

SVD. Introduction

We will present a method

for determining how close A is to amatrix of smaller rank. The method involves factoring A into aproduct UΣV T , where U is an m ×m orthogonal matrix, V is ann × n orthogonal matrix, and Σ is an m × n matrix whoseoff-diagonal entries are all 0′s and whose diagonal elements satisfy

σ1 ≥ σ2 ≥ · · · ≥ σn ≥ 0

SVD. Introduction

We will present a method for determining how close

A is to amatrix of smaller rank. The method involves factoring A into aproduct UΣV T , where U is an m ×m orthogonal matrix, V is ann × n orthogonal matrix, and Σ is an m × n matrix whoseoff-diagonal entries are all 0′s and whose diagonal elements satisfy

σ1 ≥ σ2 ≥ · · · ≥ σn ≥ 0

SVD. Introduction

We will present a method for determining how close A is

to amatrix of smaller rank. The method involves factoring A into aproduct UΣV T , where U is an m ×m orthogonal matrix, V is ann × n orthogonal matrix, and Σ is an m × n matrix whoseoff-diagonal entries are all 0′s and whose diagonal elements satisfy

σ1 ≥ σ2 ≥ · · · ≥ σn ≥ 0

SVD. Introduction

We will present a method for determining how close A is to amatrix

of smaller rank. The method involves factoring A into aproduct UΣV T , where U is an m ×m orthogonal matrix, V is ann × n orthogonal matrix, and Σ is an m × n matrix whoseoff-diagonal entries are all 0′s and whose diagonal elements satisfy

σ1 ≥ σ2 ≥ · · · ≥ σn ≥ 0

SVD. Introduction

We will present a method for determining how close A is to amatrix of smaller rank.

The method involves factoring A into aproduct UΣV T , where U is an m ×m orthogonal matrix, V is ann × n orthogonal matrix, and Σ is an m × n matrix whoseoff-diagonal entries are all 0′s and whose diagonal elements satisfy

σ1 ≥ σ2 ≥ · · · ≥ σn ≥ 0

SVD. Introduction

We will present a method for determining how close A is to amatrix of smaller rank. The method involves

factoring A into aproduct UΣV T , where U is an m ×m orthogonal matrix, V is ann × n orthogonal matrix, and Σ is an m × n matrix whoseoff-diagonal entries are all 0′s and whose diagonal elements satisfy

σ1 ≥ σ2 ≥ · · · ≥ σn ≥ 0

SVD. Introduction

We will present a method for determining how close A is to amatrix of smaller rank. The method involves factoring A

into aproduct UΣV T , where U is an m ×m orthogonal matrix, V is ann × n orthogonal matrix, and Σ is an m × n matrix whoseoff-diagonal entries are all 0′s and whose diagonal elements satisfy

σ1 ≥ σ2 ≥ · · · ≥ σn ≥ 0

SVD. Introduction

We will present a method for determining how close A is to amatrix of smaller rank. The method involves factoring A into aproduct

UΣV T , where U is an m ×m orthogonal matrix, V is ann × n orthogonal matrix, and Σ is an m × n matrix whoseoff-diagonal entries are all 0′s and whose diagonal elements satisfy

σ1 ≥ σ2 ≥ · · · ≥ σn ≥ 0

SVD. Introduction

We will present a method for determining how close A is to amatrix of smaller rank. The method involves factoring A into aproduct UΣV T ,

where U is an m ×m orthogonal matrix, V is ann × n orthogonal matrix, and Σ is an m × n matrix whoseoff-diagonal entries are all 0′s and whose diagonal elements satisfy

σ1 ≥ σ2 ≥ · · · ≥ σn ≥ 0

SVD. Introduction

We will present a method for determining how close A is to amatrix of smaller rank. The method involves factoring A into aproduct UΣV T , where U is

an m ×m orthogonal matrix, V is ann × n orthogonal matrix, and Σ is an m × n matrix whoseoff-diagonal entries are all 0′s and whose diagonal elements satisfy

σ1 ≥ σ2 ≥ · · · ≥ σn ≥ 0

SVD. Introduction

We will present a method for determining how close A is to amatrix of smaller rank. The method involves factoring A into aproduct UΣV T , where U is an m ×m

orthogonal matrix, V is ann × n orthogonal matrix, and Σ is an m × n matrix whoseoff-diagonal entries are all 0′s and whose diagonal elements satisfy

σ1 ≥ σ2 ≥ · · · ≥ σn ≥ 0

SVD. Introduction

We will present a method for determining how close A is to amatrix of smaller rank. The method involves factoring A into aproduct UΣV T , where U is an m ×m orthogonal matrix,

V is ann × n orthogonal matrix, and Σ is an m × n matrix whoseoff-diagonal entries are all 0′s and whose diagonal elements satisfy

σ1 ≥ σ2 ≥ · · · ≥ σn ≥ 0

SVD. Introduction

We will present a method for determining how close A is to amatrix of smaller rank. The method involves factoring A into aproduct UΣV T , where U is an m ×m orthogonal matrix, V is an

n × n orthogonal matrix, and Σ is an m × n matrix whoseoff-diagonal entries are all 0′s and whose diagonal elements satisfy

σ1 ≥ σ2 ≥ · · · ≥ σn ≥ 0

SVD. Introduction

We will present a method for determining how close A is to amatrix of smaller rank. The method involves factoring A into aproduct UΣV T , where U is an m ×m orthogonal matrix, V is ann × n

orthogonal matrix, and Σ is an m × n matrix whoseoff-diagonal entries are all 0′s and whose diagonal elements satisfy

σ1 ≥ σ2 ≥ · · · ≥ σn ≥ 0

SVD. Introduction

We will present a method for determining how close A is to amatrix of smaller rank. The method involves factoring A into aproduct UΣV T , where U is an m ×m orthogonal matrix, V is ann × n orthogonal matrix, and

Σ is an m × n matrix whoseoff-diagonal entries are all 0′s and whose diagonal elements satisfy

σ1 ≥ σ2 ≥ · · · ≥ σn ≥ 0

SVD. Introduction

We will present a method for determining how close A is to amatrix of smaller rank. The method involves factoring A into aproduct UΣV T , where U is an m ×m orthogonal matrix, V is ann × n orthogonal matrix, and Σ is

an m × n matrix whoseoff-diagonal entries are all 0′s and whose diagonal elements satisfy

σ1 ≥ σ2 ≥ · · · ≥ σn ≥ 0

SVD. Introduction

We will present a method for determining how close A is to amatrix of smaller rank. The method involves factoring A into aproduct UΣV T , where U is an m ×m orthogonal matrix, V is ann × n orthogonal matrix, and Σ is an m × n matrix

whoseoff-diagonal entries are all 0′s and whose diagonal elements satisfy

σ1 ≥ σ2 ≥ · · · ≥ σn ≥ 0

SVD. Introduction

We will present a method for determining how close A is to amatrix of smaller rank. The method involves factoring A into aproduct UΣV T , where U is an m ×m orthogonal matrix, V is ann × n orthogonal matrix, and Σ is an m × n matrix whoseoff-diagonal entries

are all 0′s and whose diagonal elements satisfy

σ1 ≥ σ2 ≥ · · · ≥ σn ≥ 0

SVD. Introduction

We will present a method for determining how close A is to amatrix of smaller rank. The method involves factoring A into aproduct UΣV T , where U is an m ×m orthogonal matrix, V is ann × n orthogonal matrix, and Σ is an m × n matrix whoseoff-diagonal entries are all 0′s and

whose diagonal elements satisfy

σ1 ≥ σ2 ≥ · · · ≥ σn ≥ 0

SVD. Introduction

We will present a method for determining how close A is to amatrix of smaller rank. The method involves factoring A into aproduct UΣV T , where U is an m ×m orthogonal matrix, V is ann × n orthogonal matrix, and Σ is an m × n matrix whoseoff-diagonal entries are all 0′s and whose diagonal

elements satisfy

σ1 ≥ σ2 ≥ · · · ≥ σn ≥ 0

SVD. Introduction

σ1 ≥ σ2 ≥ · · · ≥ σn ≥ 0

SVD. Introduction

σ1 ≥ σ2 ≥ · · · ≥ σn ≥ 0

SVD. Introduction

σ1σ2 . . .

The σ′s determined by this factorization are unique and are calledthe singular values of A. The factorization UΣV T is called thesingular value decomposition of A, or, for short, the SVD of A.

SVD. Introduction

σ1σ2 . . .

SVD. Introduction

σ1σ2 . . .

SVD. Introduction

σ1σ2 . . .

The σ′s determined

by this factorization are unique and are calledthe singular values of A. The factorization UΣV T is called thesingular value decomposition of A, or, for short, the SVD of A.

SVD. Introduction

σ1σ2 . . .

The σ′s determined by this factorization

are unique and are calledthe singular values of A. The factorization UΣV T is called thesingular value decomposition of A, or, for short, the SVD of A.

SVD. Introduction

σ1σ2 . . .

The σ′s determined by this factorization are unique and

are calledthe singular values of A. The factorization UΣV T is called thesingular value decomposition of A, or, for short, the SVD of A.

SVD. Introduction

σ1σ2 . . .

The σ′s determined by this factorization are unique and are called

the singular values of A. The factorization UΣV T is called thesingular value decomposition of A, or, for short, the SVD of A.

SVD. Introduction

σ1σ2 . . .

The σ′s determined by this factorization are unique and are calledthe singular values

of A. The factorization UΣV T is called thesingular value decomposition of A, or, for short, the SVD of A.

SVD. Introduction

σ1σ2 . . .

The σ′s determined by this factorization are unique and are calledthe singular values of A. The factorization

UΣV T is called thesingular value decomposition of A, or, for short, the SVD of A.

SVD. Introduction

σ1σ2 . . .

The σ′s determined by this factorization are unique and are calledthe singular values of A. The factorization UΣV T

is called thesingular value decomposition of A, or, for short, the SVD of A.

SVD. Introduction

σ1σ2 . . .

The σ′s determined by this factorization are unique and are calledthe singular values of A. The factorization UΣV T is called

thesingular value decomposition of A, or, for short, the SVD of A.

SVD. Introduction

σ1σ2 . . .

The σ′s determined by this factorization are unique and are calledthe singular values of A. The factorization UΣV T is called thesingular value decomposition

of A, or, for short, the SVD of A.

SVD. Introduction

σ1σ2 . . .

The σ′s determined by this factorization are unique and are calledthe singular values of A. The factorization UΣV T is called thesingular value decomposition of A, or,

for short, the SVD of A.

SVD. Introduction

σ1σ2 . . .

The σ′s determined by this factorization are unique and are calledthe singular values of A. The factorization UΣV T is called thesingular value decomposition of A, or, for short,

the SVD of A.

SVD. Introduction

σ1σ2 . . .

SVD. Introduction

The SVD Theorem

If A is an m × n matrix, then A has a singular value decomposition

Sketch of the proof

ATA is a symmetric n × n matrix.

The eigenvalues of ATA are all real and it has an orthogonaldiagonalizing matrix V .

Furthermore, its eigenvalues must all be nonnegative.

To see this point, let λ be an eigenvalue of ATA and x be aneigenvector belonging to λ. It follows that

SVD. Introduction

The SVD Theorem

Sketch of the proof

SVD. Introduction

The SVD Theorem

is an m × n matrix, then A has a singular value decomposition

Sketch of the proof

SVD. Introduction

The SVD Theorem

If A is an

m × n matrix, then A has a singular value decomposition

Sketch of the proof

SVD. Introduction

The SVD Theorem

If A is an m × n matrix, then

A has a singular value decomposition

Sketch of the proof

SVD. Introduction

The SVD Theorem

If A is an m × n matrix, then A has

a singular value decomposition

Sketch of the proof

SVD. Introduction

The SVD Theorem

Sketch of the proof

SVD. Introduction

The SVD Theorem

Sketch of the proof

SVD. Introduction

The SVD Theorem

Sketch of the proof

is a symmetric n × n matrix.

SVD. Introduction

The SVD Theorem

Sketch of the proof

ATA is a symmetric

n × n matrix.

SVD. Introduction

The SVD Theorem

Sketch of the proof

SVD. Introduction

The SVD Theorem

Sketch of the proof

The eigenvalues

of ATA are all real and it has an orthogonaldiagonalizing matrix V .

SVD. Introduction

The SVD Theorem

Sketch of the proof

The eigenvalues of ATA

are all real and it has an orthogonaldiagonalizing matrix V .

SVD. Introduction

The SVD Theorem

Sketch of the proof

The eigenvalues of ATA are all real and

it has an orthogonaldiagonalizing matrix V .

SVD. Introduction

The SVD Theorem

Sketch of the proof

The eigenvalues of ATA are all real and it has

an orthogonaldiagonalizing matrix V .

SVD. Introduction

The SVD Theorem

Sketch of the proof

The eigenvalues of ATA are all real and it has an orthogonal

diagonalizing matrix V .

SVD. Introduction

The SVD Theorem

Sketch of the proof

The eigenvalues of ATA are all real and it has an orthogonaldiagonalizing matrix

SVD. Introduction

The SVD Theorem

Sketch of the proof

SVD. Introduction

The SVD Theorem

Sketch of the proof

Furthermore,

its eigenvalues must all be nonnegative.

SVD. Introduction

The SVD Theorem

Sketch of the proof

Furthermore, its eigenvalues

must all be nonnegative.

SVD. Introduction

The SVD Theorem

Sketch of the proof

Furthermore, its eigenvalues must all be

nonnegative.

SVD. Introduction

The SVD Theorem

Sketch of the proof

SVD. Introduction

The SVD Theorem

Sketch of the proof

To see

this point, let λ be an eigenvalue of ATA and x be aneigenvector belonging to λ. It follows that

SVD. Introduction

The SVD Theorem

Sketch of the proof

To see this point,

let λ be an eigenvalue of ATA and x be aneigenvector belonging to λ. It follows that

SVD. Introduction

The SVD Theorem

Sketch of the proof

To see this point, let λ

be an eigenvalue of ATA and x be aneigenvector belonging to λ. It follows that

SVD. Introduction

The SVD Theorem

Sketch of the proof

To see this point, let λ be an eigenvalue

of ATA and x be aneigenvector belonging to λ. It follows that

SVD. Introduction

The SVD Theorem

Sketch of the proof

To see this point, let λ be an eigenvalue of ATA and

x be aneigenvector belonging to λ. It follows that

SVD. Introduction

The SVD Theorem

Sketch of the proof

To see this point, let λ be an eigenvalue of ATA and x

be aneigenvector belonging to λ. It follows that

SVD. Introduction

The SVD Theorem

Sketch of the proof

To see this point, let λ be an eigenvalue of ATA and x be aneigenvector

belonging to λ. It follows that

SVD. Introduction

The SVD Theorem

Sketch of the proof

To see this point, let λ be an eigenvalue of ATA and x be aneigenvector belonging to λ.

It follows that

SVD. Introduction

The SVD Theorem

Sketch of the proof

SVD. Introduction

||Ax||2 = xTATAx = xTλx = λxTx = λ||x||2 ⇒

λ =||Ax||2

||x||2

We may assume that the columns of V have been ordered sothat the corresponding eigenvalues satisfyλ1 ≥ λ2 ≥ · · · ≥ λn ≥ 0. The singular values are given by

σj =√λj , j = 1, 2, ..., n

SVD. Introduction

||Ax||2 =

xTATAx = xTλx = λxTx = λ||x||2 ⇒

λ =||Ax||2

||x||2

σj =√λj , j = 1, 2, ..., n

SVD. Introduction

||Ax||2 = xTATAx =

xTλx = λxTx = λ||x||2 ⇒

λ =||Ax||2

||x||2

σj =√λj , j = 1, 2, ..., n

SVD. Introduction

||Ax||2 = xTATAx = xTλx =

λxTx = λ||x||2 ⇒

λ =||Ax||2

||x||2

σj =√λj , j = 1, 2, ..., n

SVD. Introduction

λ =||Ax||2

||x||2

σj =√λj , j = 1, 2, ..., n

SVD. Introduction

λ =||Ax||2

||x||2

σj =√λj , j = 1, 2, ..., n

SVD. Introduction

||Ax||2

||x||2

σj =√λj , j = 1, 2, ..., n

SVD. Introduction

λ =||Ax||2

||x||2

σj =√λj , j = 1, 2, ..., n

SVD. Introduction

λ =||Ax||2

||x||2

We may assume

that the columns of V have been ordered sothat the corresponding eigenvalues satisfyλ1 ≥ λ2 ≥ · · · ≥ λn ≥ 0. The singular values are given by

σj =√λj , j = 1, 2, ..., n

SVD. Introduction

λ =||Ax||2

||x||2

We may assume that the columns

of V have been ordered sothat the corresponding eigenvalues satisfyλ1 ≥ λ2 ≥ · · · ≥ λn ≥ 0. The singular values are given by

σj =√λj , j = 1, 2, ..., n

SVD. Introduction

λ =||Ax||2

||x||2

We may assume that the columns of V

have been ordered sothat the corresponding eigenvalues satisfyλ1 ≥ λ2 ≥ · · · ≥ λn ≥ 0. The singular values are given by

σj =√λj , j = 1, 2, ..., n

SVD. Introduction

λ =||Ax||2

||x||2

We may assume that the columns of V have been ordered

sothat the corresponding eigenvalues satisfyλ1 ≥ λ2 ≥ · · · ≥ λn ≥ 0. The singular values are given by

σj =√λj , j = 1, 2, ..., n

SVD. Introduction

λ =||Ax||2

||x||2

We may assume that the columns of V have been ordered sothat

the corresponding eigenvalues satisfyλ1 ≥ λ2 ≥ · · · ≥ λn ≥ 0. The singular values are given by

σj =√λj , j = 1, 2, ..., n

SVD. Introduction

λ =||Ax||2

||x||2

We may assume that the columns of V have been ordered sothat the corresponding

eigenvalues satisfyλ1 ≥ λ2 ≥ · · · ≥ λn ≥ 0. The singular values are given by

σj =√λj , j = 1, 2, ..., n

SVD. Introduction

λ =||Ax||2

||x||2

We may assume that the columns of V have been ordered sothat the corresponding eigenvalues satisfy

λ1 ≥ λ2 ≥ · · · ≥ λn ≥ 0. The singular values are given by

σj =√λj , j = 1, 2, ..., n

SVD. Introduction

λ =||Ax||2

||x||2

We may assume that the columns of V have been ordered sothat the corresponding eigenvalues satisfyλ1 ≥ λ2 ≥ · · · ≥ λn ≥ 0.

The singular values are given by

σj =√λj , j = 1, 2, ..., n

SVD. Introduction

λ =||Ax||2

||x||2

We may assume that the columns of V have been ordered sothat the corresponding eigenvalues satisfyλ1 ≥ λ2 ≥ · · · ≥ λn ≥ 0. The singular values

are given by

σj =√λj , j = 1, 2, ..., n

SVD. Introduction

λ =||Ax||2

||x||2

σj =√λj , j = 1, 2, ..., n

SVD. Introduction

λ =||Ax||2

||x||2

√λj , j = 1, 2, ..., n

SVD. Introduction

λ =||Ax||2

||x||2

σj =√λj ,

j = 1, 2, ..., n

SVD. Introduction

λ =||Ax||2

||x||2

σj =√λj , j = 1, 2, ..., n

SVD. Introduction

Let r denote the rank of A. The matrix ATA will also haverank r . Since ATA is symmetric, its rank equals the numberof nonzero eigenvalues. Thus,

σ1 ≥ σ2 ≥ · · · ≥ σr > 0 σr+1 = σr+2 = · · · = σn = 0

Now let V1 = (v1, v2, ...., vr , ) and V2 = (vr+1, vr+2, ...., vn, )

The column vectors of V1 are eigenvectors of ATA belongingto λi , i = 1, 2, ..., r .

The column vectors of V2 are eigenvectors of ATA belongingto λj = 0, j = r + 1, r + 2, ..., n.

SVD. Introduction

Let r denote

the rank of A. The matrix ATA will also haverank r . Since ATA is symmetric, its rank equals the numberof nonzero eigenvalues. Thus,

σ1 ≥ σ2 ≥ · · · ≥ σr > 0 σr+1 = σr+2 = · · · = σn = 0

SVD. Introduction

Let r denote the rank

of A. The matrix ATA will also haverank r . Since ATA is symmetric, its rank equals the numberof nonzero eigenvalues. Thus,

σ1 ≥ σ2 ≥ · · · ≥ σr > 0 σr+1 = σr+2 = · · · = σn = 0

SVD. Introduction

Let r denote the rank of A.

The matrix ATA will also haverank r . Since ATA is symmetric, its rank equals the numberof nonzero eigenvalues. Thus,

σ1 ≥ σ2 ≥ · · · ≥ σr > 0 σr+1 = σr+2 = · · · = σn = 0

SVD. Introduction

Let r denote the rank of A. The matrix

ATA will also haverank r . Since ATA is symmetric, its rank equals the numberof nonzero eigenvalues. Thus,

σ1 ≥ σ2 ≥ · · · ≥ σr > 0 σr+1 = σr+2 = · · · = σn = 0

SVD. Introduction

Let r denote the rank of A. The matrix ATA

will also haverank r . Since ATA is symmetric, its rank equals the numberof nonzero eigenvalues. Thus,

σ1 ≥ σ2 ≥ · · · ≥ σr > 0 σr+1 = σr+2 = · · · = σn = 0

SVD. Introduction

Let r denote the rank of A. The matrix ATA will also have

rank r . Since ATA is symmetric, its rank equals the numberof nonzero eigenvalues. Thus,

σ1 ≥ σ2 ≥ · · · ≥ σr > 0 σr+1 = σr+2 = · · · = σn = 0

SVD. Introduction

Let r denote the rank of A. The matrix ATA will also haverank r .

Since ATA is symmetric, its rank equals the numberof nonzero eigenvalues. Thus,

σ1 ≥ σ2 ≥ · · · ≥ σr > 0 σr+1 = σr+2 = · · · = σn = 0

SVD. Introduction

Let r denote the rank of A. The matrix ATA will also haverank r . Since ATA

is symmetric, its rank equals the numberof nonzero eigenvalues. Thus,

σ1 ≥ σ2 ≥ · · · ≥ σr > 0 σr+1 = σr+2 = · · · = σn = 0

SVD. Introduction

Let r denote the rank of A. The matrix ATA will also haverank r . Since ATA is symmetric,

its rank equals the numberof nonzero eigenvalues. Thus,

σ1 ≥ σ2 ≥ · · · ≥ σr > 0 σr+1 = σr+2 = · · · = σn = 0

SVD. Introduction

Let r denote the rank of A. The matrix ATA will also haverank r . Since ATA is symmetric, its rank equals

the numberof nonzero eigenvalues. Thus,

σ1 ≥ σ2 ≥ · · · ≥ σr > 0 σr+1 = σr+2 = · · · = σn = 0

SVD. Introduction

Let r denote the rank of A. The matrix ATA will also haverank r . Since ATA is symmetric, its rank equals the numberof

nonzero eigenvalues. Thus,

σ1 ≥ σ2 ≥ · · · ≥ σr > 0 σr+1 = σr+2 = · · · = σn = 0

SVD. Introduction

Let r denote the rank of A. The matrix ATA will also haverank r . Since ATA is symmetric, its rank equals the numberof nonzero eigenvalues.

σ1 ≥ σ2 ≥ · · · ≥ σr > 0 σr+1 = σr+2 = · · · = σn = 0

SVD. Introduction

σ1 ≥ σ2 ≥ · · · ≥ σr > 0 σr+1 = σr+2 = · · · = σn = 0

SVD. Introduction

σ1 ≥ σ2 ≥ · · · ≥ σr > 0

σr+1 = σr+2 = · · · = σn = 0

SVD. Introduction

σ1 ≥ σ2 ≥ · · · ≥ σr > 0 σr+1 = σr+2 = · · · = σn = 0

SVD. Introduction

σ1 ≥ σ2 ≥ · · · ≥ σr > 0 σr+1 = σr+2 = · · · = σn = 0

let V1 = (v1, v2, ...., vr , ) and V2 = (vr+1, vr+2, ...., vn, )

SVD. Introduction

σ1 ≥ σ2 ≥ · · · ≥ σr > 0 σr+1 = σr+2 = · · · = σn = 0

Now let V1 = (v1, v2, ...., vr , ) and

V2 = (vr+1, vr+2, ...., vn, )

SVD. Introduction

σ1 ≥ σ2 ≥ · · · ≥ σr > 0 σr+1 = σr+2 = · · · = σn = 0

SVD. Introduction

σ1 ≥ σ2 ≥ · · · ≥ σr > 0 σr+1 = σr+2 = · · · = σn = 0

The column vectors

of V1 are eigenvectors of ATA belongingto λi , i = 1, 2, ..., r .

SVD. Introduction

σ1 ≥ σ2 ≥ · · · ≥ σr > 0 σr+1 = σr+2 = · · · = σn = 0

The column vectors of V1

are eigenvectors of ATA belongingto λi , i = 1, 2, ..., r .

SVD. Introduction

σ1 ≥ σ2 ≥ · · · ≥ σr > 0 σr+1 = σr+2 = · · · = σn = 0

The column vectors of V1 are eigenvectors of

ATA belongingto λi , i = 1, 2, ..., r .

SVD. Introduction

σ1 ≥ σ2 ≥ · · · ≥ σr > 0 σr+1 = σr+2 = · · · = σn = 0

The column vectors of V1 are eigenvectors of ATA

belongingto λi , i = 1, 2, ..., r .

SVD. Introduction

σ1 ≥ σ2 ≥ · · · ≥ σr > 0 σr+1 = σr+2 = · · · = σn = 0

The column vectors of V1 are eigenvectors of ATA belongingto

λi , i = 1, 2, ..., r .

SVD. Introduction

σ1 ≥ σ2 ≥ · · · ≥ σr > 0 σr+1 = σr+2 = · · · = σn = 0

SVD. Introduction

σ1 ≥ σ2 ≥ · · · ≥ σr > 0 σr+1 = σr+2 = · · · = σn = 0

The column vectors

of V2 are eigenvectors of ATA belongingto λj = 0, j = r + 1, r + 2, ..., n.

SVD. Introduction

σ1 ≥ σ2 ≥ · · · ≥ σr > 0 σr+1 = σr+2 = · · · = σn = 0

The column vectors of V2

are eigenvectors of ATA belongingto λj = 0, j = r + 1, r + 2, ..., n.

SVD. Introduction

σ1 ≥ σ2 ≥ · · · ≥ σr > 0 σr+1 = σr+2 = · · · = σn = 0

The column vectors of V2 are eigenvectors

of ATA belongingto λj = 0, j = r + 1, r + 2, ..., n.

SVD. Introduction

σ1 ≥ σ2 ≥ · · · ≥ σr > 0 σr+1 = σr+2 = · · · = σn = 0

The column vectors of V2 are eigenvectors of ATA

belongingto λj = 0, j = r + 1, r + 2, ..., n.

SVD. Introduction

σ1 ≥ σ2 ≥ · · · ≥ σr > 0 σr+1 = σr+2 = · · · = σn = 0

The column vectors of V2 are eigenvectors of ATA belongingto

λj = 0, j = r + 1, r + 2, ..., n.

SVD. Introduction

σ1 ≥ σ2 ≥ · · · ≥ σr > 0 σr+1 = σr+2 = · · · = σn = 0

SVD. Introduction

Now let Σ1 be the r × r matrix defined by

σ1σ2 . . .

The m × n matrix Σ is then given by

(Σ1 00 0

SVD. Introduction

let Σ1 be the r × r matrix defined by

σ1σ2 . . .

(Σ1 00 0

SVD. Introduction

Now let Σ1

be the r × r matrix defined by

σ1σ2 . . .

(Σ1 00 0

SVD. Introduction

Now let Σ1 be the r × r matrix

defined by

σ1σ2 . . .

(Σ1 00 0

SVD. Introduction

σ1σ2 . . .

(Σ1 00 0

SVD. Introduction

σ1σ2 . . .

(Σ1 00 0

SVD. Introduction

σ1σ2 . . .

(Σ1 00 0

SVD. Introduction

σ1σ2 . . .

The m × n matrix

Σ is then given by

(Σ1 00 0

SVD. Introduction

σ1σ2 . . .

The m × n matrix Σ

is then given by

(Σ1 00 0

SVD. Introduction

σ1σ2 . . .

(Σ1 00 0

SVD. Introduction

σ1σ2 . . .

(Σ1 00 0

SVD. Introduction

σ1σ2 . . .

(Σ1 00 0

SVD. Introduction

To complete the proof, we must show how to construct anm ×m orthogonal matrix U such that

A = UΣV T

AV = UΣ

Comparing the first r columns of each side of the lastequation, we see that

Avi = σivi , i = 1, 2, ..., r

Thus, if we define

σiAvi , i = 1, 2, ..., r

SVD. Introduction

A = UΣV T

AV = UΣ

Avi = σivi , i = 1, 2, ..., r

Thus, if we define

σiAvi , i = 1, 2, ..., r

SVD. Introduction

A = UΣV T

AV = UΣ

Avi = σivi , i = 1, 2, ..., r

Thus, if we define

σiAvi , i = 1, 2, ..., r

SVD. Introduction

A = UΣV T

AV = UΣ

Avi = σivi , i = 1, 2, ..., r

Thus, if we define

σiAvi , i = 1, 2, ..., r

SVD. Introduction

A = UΣV T

AV = UΣ

Avi = σivi , i = 1, 2, ..., r

Thus, if we define

σiAvi , i = 1, 2, ..., r

SVD. Introduction

A = UΣV T

AV = UΣ

Avi = σivi , i = 1, 2, ..., r

Thus, if we define

σiAvi , i = 1, 2, ..., r

SVD. Introduction

A = UΣV T

AV = UΣ

Avi = σivi , i = 1, 2, ..., r

Thus, if we define

σiAvi , i = 1, 2, ..., r

SVD. Introduction

A = UΣV T

AV = UΣ

Avi = σivi , i = 1, 2, ..., r

Thus, if we define

σiAvi , i = 1, 2, ..., r

SVD. Introduction

U1 = (u1,u2, ...,ur )

then it follows that

AV1 = U1Σ1

The column vectors of U1 form an orthonormal set. Thus,form an orthonormal basis for R(A). The vector spaceR(A)⊥ = N(AT ) has dimension m − r . Let{ur+1,ur+2, · · · ,un} be an orthonormal basis for N(AT ) andset

U2 = (ur+1,ur+2, ...,un)

SVD. Introduction

U1 = (u1,u2, ...,ur )

AV1 = U1Σ1

U2 = (ur+1,ur+2, ...,un)

SVD. Introduction

U1 = (u1,u2, ...,ur )

AV1 = U1Σ1

U2 = (ur+1,ur+2, ...,un)

SVD. Introduction

U1 = (u1,u2, ...,ur )

AV1 = U1Σ1

U2 = (ur+1,ur+2, ...,un)

SVD. Introduction

U1 = (u1,u2, ...,ur )

AV1 = U1Σ1

U2 = (ur+1,ur+2, ...,un)

SVD. Introduction

U1 = (u1,u2, ...,ur )

AV1 = U1Σ1

The column vectors

of U1 form an orthonormal set. Thus,form an orthonormal basis for R(A). The vector spaceR(A)⊥ = N(AT ) has dimension m − r . Let{ur+1,ur+2, · · · ,un} be an orthonormal basis for N(AT ) andset

U2 = (ur+1,ur+2, ...,un)

SVD. Introduction

U1 = (u1,u2, ...,ur )

AV1 = U1Σ1

The column vectors of U1

form an orthonormal set. Thus,form an orthonormal basis for R(A). The vector spaceR(A)⊥ = N(AT ) has dimension m − r . Let{ur+1,ur+2, · · · ,un} be an orthonormal basis for N(AT ) andset

U2 = (ur+1,ur+2, ...,un)

SVD. Introduction

U1 = (u1,u2, ...,ur )

AV1 = U1Σ1

The column vectors of U1 form an orthonormal set.

Thus,form an orthonormal basis for R(A). The vector spaceR(A)⊥ = N(AT ) has dimension m − r . Let{ur+1,ur+2, · · · ,un} be an orthonormal basis for N(AT ) andset

U2 = (ur+1,ur+2, ...,un)

SVD. Introduction

U1 = (u1,u2, ...,ur )

AV1 = U1Σ1

The column vectors of U1 form an orthonormal set. Thus,

form an orthonormal basis for R(A). The vector spaceR(A)⊥ = N(AT ) has dimension m − r . Let{ur+1,ur+2, · · · ,un} be an orthonormal basis for N(AT ) andset

U2 = (ur+1,ur+2, ...,un)

SVD. Introduction

U1 = (u1,u2, ...,ur )

AV1 = U1Σ1

The column vectors of U1 form an orthonormal set. Thus,form an orthonormal

basis for R(A). The vector spaceR(A)⊥ = N(AT ) has dimension m − r . Let{ur+1,ur+2, · · · ,un} be an orthonormal basis for N(AT ) andset

U2 = (ur+1,ur+2, ...,un)

SVD. Introduction

U1 = (u1,u2, ...,ur )

AV1 = U1Σ1

The column vectors of U1 form an orthonormal set. Thus,form an orthonormal basis for

R(A). The vector spaceR(A)⊥ = N(AT ) has dimension m − r . Let{ur+1,ur+2, · · · ,un} be an orthonormal basis for N(AT ) andset

U2 = (ur+1,ur+2, ...,un)

SVD. Introduction

U1 = (u1,u2, ...,ur )

AV1 = U1Σ1

The column vectors of U1 form an orthonormal set. Thus,form an orthonormal basis for R(A).

The vector spaceR(A)⊥ = N(AT ) has dimension m − r . Let{ur+1,ur+2, · · · ,un} be an orthonormal basis for N(AT ) andset

U2 = (ur+1,ur+2, ...,un)

SVD. Introduction

U1 = (u1,u2, ...,ur )

AV1 = U1Σ1

The column vectors of U1 form an orthonormal set. Thus,form an orthonormal basis for R(A). The vector space

R(A)⊥ = N(AT ) has dimension m − r . Let{ur+1,ur+2, · · · ,un} be an orthonormal basis for N(AT ) andset

U2 = (ur+1,ur+2, ...,un)

SVD. Introduction

U1 = (u1,u2, ...,ur )

AV1 = U1Σ1

The column vectors of U1 form an orthonormal set. Thus,form an orthonormal basis for R(A). The vector spaceR(A)⊥ = N(AT )

has dimension m − r . Let{ur+1,ur+2, · · · ,un} be an orthonormal basis for N(AT ) andset

U2 = (ur+1,ur+2, ...,un)

SVD. Introduction

U1 = (u1,u2, ...,ur )

AV1 = U1Σ1

The column vectors of U1 form an orthonormal set. Thus,form an orthonormal basis for R(A). The vector spaceR(A)⊥ = N(AT ) has dimension

m − r . Let{ur+1,ur+2, · · · ,un} be an orthonormal basis for N(AT ) andset

U2 = (ur+1,ur+2, ...,un)

SVD. Introduction

U1 = (u1,u2, ...,ur )

AV1 = U1Σ1

The column vectors of U1 form an orthonormal set. Thus,form an orthonormal basis for R(A). The vector spaceR(A)⊥ = N(AT ) has dimension m − r .

Let{ur+1,ur+2, · · · ,un} be an orthonormal basis for N(AT ) andset

U2 = (ur+1,ur+2, ...,un)

SVD. Introduction

U1 = (u1,u2, ...,ur )

AV1 = U1Σ1

The column vectors of U1 form an orthonormal set. Thus,form an orthonormal basis for R(A). The vector spaceR(A)⊥ = N(AT ) has dimension m − r . Let

{ur+1,ur+2, · · · ,un} be an orthonormal basis for N(AT ) andset

U2 = (ur+1,ur+2, ...,un)

SVD. Introduction

U1 = (u1,u2, ...,ur )

AV1 = U1Σ1

The column vectors of U1 form an orthonormal set. Thus,form an orthonormal basis for R(A). The vector spaceR(A)⊥ = N(AT ) has dimension m − r . Let{ur+1,ur+2, · · · ,un}

be an orthonormal basis for N(AT ) andset

U2 = (ur+1,ur+2, ...,un)

SVD. Introduction

U1 = (u1,u2, ...,ur )

AV1 = U1Σ1

The column vectors of U1 form an orthonormal set. Thus,form an orthonormal basis for R(A). The vector spaceR(A)⊥ = N(AT ) has dimension m − r . Let{ur+1,ur+2, · · · ,un} be an orthonormal basis

for N(AT ) andset

U2 = (ur+1,ur+2, ...,un)

SVD. Introduction

U1 = (u1,u2, ...,ur )

AV1 = U1Σ1

U2 = (ur+1,ur+2, ...,un)

SVD. Introduction

U1 = (u1,u2, ...,ur )

AV1 = U1Σ1

U2 = (ur+1,ur+2, ...,un)

SVD. Introduction

Setthe m × n matrix U by

U = (U1,U2)

The the matrices U,Σ, and V satisfy

A = UΣV T �

SVD. Introduction

U = (U1,U2)

A = UΣV T �

SVD. Introduction

U = (U1,U2)

A = UΣV T �

SVD. Introduction

U = (U1,U2)

A = UΣV T �

SVD. Introduction

U = (U1,U2)

UΣV T �

SVD. Introduction

U = (U1,U2)

ΣV T �

SVD. Introduction

U = (U1,U2)

A = UΣ

V T �

SVD. Introduction

U = (U1,U2)

A = UΣV T

SVD. Introduction

U = (U1,U2)

A = UΣV T �

SVD. Introduction

U = (U1,U2)

A = UΣV T �

SVD. Introduction

Observations

Let A be an m × n matrix with a singular value decompositionA = UΣV T .

The singular values σ1, ..., σn of A are unique; however, thematrices U and V are not unique.

Since V diagonalizes ATA, it follows that the v′js are

eigenvectors of ATA.

Since AAT = UΣΣTUT , it follows that U diagonalizes AAT

and that the u′js are eigenvectors of AAT .

The v′js are called the right singular vectors of A, and the u′jsare called the left singular vectors of A.

SVD. Introduction

Observations

SVD. Introduction

Observations

SVD. Introduction

Observations

SVD. Introduction

Observations

SVD. Introduction

Observations

SVD. Introduction

Observations

SVD. Introduction

Observations

SVD. Introduction

Observations

SVD. Introduction

Observations

SVD. Introduction

If A has rank r , then

(i) v1, v2, ..., vr form an orthonormal basis for R(AT ).

(ii) vr+1, vr+2, ..., vn form an orthonormal basis for N(A).

(iii) u1,u2, ...,ur form an orthonormal basis for R(A).

(iv) ur+1,ur+2, ...,ur+n form an orthonormal basis for N(AT )

The rank of the matrix A is equal to the number of itsnonzero singular values (where singular values are countedaccording to multiplicity).

SVD. Introduction

In the case that A has rank r < n, if we set

V1 = (v1, v2, ..., vr ) U1 = (u1,u2, ...,ur )and define Σ1 as before, then

A = U1Σ1VT1

This factorization, is called the compact form of thesingular value decomposition of A. This form is useful inmany applications

SVD. Introduction

A = U1Σ1VT1

SVD. Introduction

V1 = (v1, v2, ..., vr )

U1 = (u1,u2, ...,ur )and define Σ1 as before, then

A = U1Σ1VT1

SVD. Introduction

A = U1Σ1VT1

SVD. Introduction

U1Σ1VT1

SVD. Introduction

A = U1

Σ1VT1

SVD. Introduction

A = U1Σ1

SVD. Introduction

A = U1Σ1VT1

SVD. Introduction

A = U1Σ1VT1

This factorization,

is called the compact form of thesingular value decomposition of A. This form is useful inmany applications

SVD. Introduction

A = U1Σ1VT1

This factorization, is called

the compact form of thesingular value decomposition of A. This form is useful inmany applications

SVD. Introduction

A = U1Σ1VT1

This factorization, is called the compact form of thesingular value decomposition

of A. This form is useful inmany applications

SVD. Introduction

A = U1Σ1VT1

This factorization, is called the compact form of thesingular value decomposition of A. This form

is useful inmany applications

SVD. Introduction

A = U1Σ1VT1

This factorization, is called the compact form of thesingular value decomposition of A. This form is useful

inmany applications

SVD. Introduction

A = U1Σ1VT1

SVD. Examples

Example 12.7let

1 11 10 0

Compute the singular values and the singular value decompositionof A

Solution

The matrix

(2 22 2

SVD. Examples

Example 12.7

1 11 10 0

Solution

The matrix

(2 22 2

SVD. Examples

Example 12.7let

1 11 10 0

Solution

The matrix

(2 22 2

SVD. Examples

Example 12.7let

1 11 10 0

Solution

The matrix

(2 22 2

SVD. Examples

Example 12.7let

1 11 10 0

Solution

The matrix

(2 22 2

SVD. Examples

Example 12.7let

1 11 10 0

Compute

the singular values and the singular value decompositionof A

Solution

The matrix

(2 22 2

SVD. Examples

Example 12.7let

1 11 10 0

Compute the singular values and

the singular value decompositionof A

Solution

The matrix

(2 22 2

SVD. Examples

Example 12.7let

1 11 10 0

Compute the singular values and the singular value decomposition

Solution

The matrix

(2 22 2

SVD. Examples

Example 12.7let

1 11 10 0

Solution

The matrix

(2 22 2

SVD. Examples

Example 12.7let

1 11 10 0

Solution

The matrix

(2 22 2

SVD. Examples

Example 12.7let

1 11 10 0

Solution

The matrix

(2 22 2

SVD. Examples

Example 12.7let

1 11 10 0

Solution

The matrix

(2 22 2

SVD. Examples

Example 12.7let

1 11 10 0

Solution

The matrix

(2 22 2

SVD. Examples

has eigenvalues λ1 = 4 and λ2 = 0.Consequently, the singularvalues of A are σ1 = 2 and σ2 = 0 The eigenvalue λ1 haseigenvectors of the form α(1, 1)T , and σ2 has eigenvectors of theform β(1, 1)T . Therefore, the orthogonal matrix

V =1√2

(1 11 −1

SVD. Examples

has eigenvalues

λ1 = 4 and λ2 = 0.Consequently, the singularvalues of A are σ1 = 2 and σ2 = 0 The eigenvalue λ1 haseigenvectors of the form α(1, 1)T , and σ2 has eigenvectors of theform β(1, 1)T . Therefore, the orthogonal matrix

V =1√2

(1 11 −1

SVD. Examples

has eigenvalues λ1 = 4 and λ2 = 0.

Consequently, the singularvalues of A are σ1 = 2 and σ2 = 0 The eigenvalue λ1 haseigenvectors of the form α(1, 1)T , and σ2 has eigenvectors of theform β(1, 1)T . Therefore, the orthogonal matrix

V =1√2

(1 11 −1

SVD. Examples

has eigenvalues λ1 = 4 and λ2 = 0.Consequently,

the singularvalues of A are σ1 = 2 and σ2 = 0 The eigenvalue λ1 haseigenvectors of the form α(1, 1)T , and σ2 has eigenvectors of theform β(1, 1)T . Therefore, the orthogonal matrix

V =1√2

(1 11 −1

SVD. Examples

has eigenvalues λ1 = 4 and λ2 = 0.Consequently, the singularvalues

of A are σ1 = 2 and σ2 = 0 The eigenvalue λ1 haseigenvectors of the form α(1, 1)T , and σ2 has eigenvectors of theform β(1, 1)T . Therefore, the orthogonal matrix

V =1√2

(1 11 −1

SVD. Examples

has eigenvalues λ1 = 4 and λ2 = 0.Consequently, the singularvalues of A are

σ1 = 2 and σ2 = 0 The eigenvalue λ1 haseigenvectors of the form α(1, 1)T , and σ2 has eigenvectors of theform β(1, 1)T . Therefore, the orthogonal matrix

V =1√2

(1 11 −1

SVD. Examples

has eigenvalues λ1 = 4 and λ2 = 0.Consequently, the singularvalues of A are σ1 = 2 and

σ2 = 0 The eigenvalue λ1 haseigenvectors of the form α(1, 1)T , and σ2 has eigenvectors of theform β(1, 1)T . Therefore, the orthogonal matrix

V =1√2

(1 11 −1

SVD. Examples

has eigenvalues λ1 = 4 and λ2 = 0.Consequently, the singularvalues of A are σ1 = 2 and σ2 = 0

The eigenvalue λ1 haseigenvectors of the form α(1, 1)T , and σ2 has eigenvectors of theform β(1, 1)T . Therefore, the orthogonal matrix

V =1√2

(1 11 −1

SVD. Examples

has eigenvalues λ1 = 4 and λ2 = 0.Consequently, the singularvalues of A are σ1 = 2 and σ2 = 0 The eigenvalue

λ1 haseigenvectors of the form α(1, 1)T , and σ2 has eigenvectors of theform β(1, 1)T . Therefore, the orthogonal matrix

V =1√2

(1 11 −1

SVD. Examples

has eigenvalues λ1 = 4 and λ2 = 0.Consequently, the singularvalues of A are σ1 = 2 and σ2 = 0 The eigenvalue λ1

haseigenvectors of the form α(1, 1)T , and σ2 has eigenvectors of theform β(1, 1)T . Therefore, the orthogonal matrix

V =1√2

(1 11 −1

SVD. Examples

has eigenvalues λ1 = 4 and λ2 = 0.Consequently, the singularvalues of A are σ1 = 2 and σ2 = 0 The eigenvalue λ1 haseigenvectors

of the form α(1, 1)T , and σ2 has eigenvectors of theform β(1, 1)T . Therefore, the orthogonal matrix

V =1√2

(1 11 −1

SVD. Examples

has eigenvalues λ1 = 4 and λ2 = 0.Consequently, the singularvalues of A are σ1 = 2 and σ2 = 0 The eigenvalue λ1 haseigenvectors of the form

α(1, 1)T , and σ2 has eigenvectors of theform β(1, 1)T . Therefore, the orthogonal matrix

V =1√2

(1 11 −1

SVD. Examples

has eigenvalues λ1 = 4 and λ2 = 0.Consequently, the singularvalues of A are σ1 = 2 and σ2 = 0 The eigenvalue λ1 haseigenvectors of the form α(1, 1)T , and

σ2 has eigenvectors of theform β(1, 1)T . Therefore, the orthogonal matrix

V =1√2

(1 11 −1

SVD. Examples

has eigenvalues λ1 = 4 and λ2 = 0.Consequently, the singularvalues of A are σ1 = 2 and σ2 = 0 The eigenvalue λ1 haseigenvectors of the form α(1, 1)T , and σ2 has eigenvectors

of theform β(1, 1)T . Therefore, the orthogonal matrix

V =1√2

(1 11 −1

SVD. Examples

has eigenvalues λ1 = 4 and λ2 = 0.Consequently, the singularvalues of A are σ1 = 2 and σ2 = 0 The eigenvalue λ1 haseigenvectors of the form α(1, 1)T , and σ2 has eigenvectors of theform

β(1, 1)T . Therefore, the orthogonal matrix

V =1√2

(1 11 −1

SVD. Examples

has eigenvalues λ1 = 4 and λ2 = 0.Consequently, the singularvalues of A are σ1 = 2 and σ2 = 0 The eigenvalue λ1 haseigenvectors of the form α(1, 1)T , and σ2 has eigenvectors of theform β(1, 1)T .

Therefore, the orthogonal matrix

V =1√2

(1 11 −1

SVD. Examples

has eigenvalues λ1 = 4 and λ2 = 0.Consequently, the singularvalues of A are σ1 = 2 and σ2 = 0 The eigenvalue λ1 haseigenvectors of the form α(1, 1)T , and σ2 has eigenvectors of theform β(1, 1)T . Therefore,

the orthogonal matrix

V =1√2

(1 11 −1

SVD. Examples

V =1√2

(1 11 −1

SVD. Examples

(1 11 −1

SVD. Examples

V =1√2

(1 11 −1

SVD. Examples

V =1√2

(1 11 −1

SVD. Examples

diagonalizes ATA. From what we discussed before, it follows that

σ1Av1 =

1 11 10 0

(1/√

The remaining column vectors of U must form an orthonormalbasis for N(AT ). We can compute a basis {x2, x3} for N(AT ) inthe usual way.

x2 = (1,−1, 0)T , x3 = (0, 0, 1)T

Since these vectors are already orthogonal, it is not necessary touse the Gram-Schmidt process to obtain an orthonormal basis.

SVD. Examples

diagonalizes

ATA. From what we discussed before, it follows that

σ1Av1 =

1 11 10 0

(1/√

x2 = (1,−1, 0)T , x3 = (0, 0, 1)T

SVD. Examples

diagonalizes ATA.

From what we discussed before, it follows that

σ1Av1 =

1 11 10 0

(1/√

x2 = (1,−1, 0)T , x3 = (0, 0, 1)T

SVD. Examples

diagonalizes ATA. From what we discussed before,

it follows that

σ1Av1 =

1 11 10 0

(1/√

x2 = (1,−1, 0)T , x3 = (0, 0, 1)T

SVD. Examples

σ1Av1 =

1 11 10 0

(1/√

x2 = (1,−1, 0)T , x3 = (0, 0, 1)T

SVD. Examples

σ1Av1 =

1 11 10 0

(1/√

x2 = (1,−1, 0)T , x3 = (0, 0, 1)T

SVD. Examples

σ1Av1 =

1 11 10 0

(1/√

x2 = (1,−1, 0)T , x3 = (0, 0, 1)T

SVD. Examples

σ1Av1 =

1 11 10 0

(1/√

x2 = (1,−1, 0)T , x3 = (0, 0, 1)T

SVD. Examples

σ1Av1 =

1 11 10 0

(1/√

x2 = (1,−1, 0)T , x3 = (0, 0, 1)T

SVD. Examples

σ1Av1 =

1 11 10 0

(1/√

x2 = (1,−1, 0)T , x3 = (0, 0, 1)T

SVD. Examples

σ1Av1 =

1 11 10 0

(1/√

x2 = (1,−1, 0)T , x3 = (0, 0, 1)T

SVD. Examples

σ1Av1 =

1 11 10 0

(1/√

The remaining

column vectors of U must form an orthonormalbasis for N(AT ). We can compute a basis {x2, x3} for N(AT ) inthe usual way.

x2 = (1,−1, 0)T , x3 = (0, 0, 1)T

SVD. Examples

σ1Av1 =

1 11 10 0

(1/√

The remaining column vectors

of U must form an orthonormalbasis for N(AT ). We can compute a basis {x2, x3} for N(AT ) inthe usual way.

x2 = (1,−1, 0)T , x3 = (0, 0, 1)T

SVD. Examples

σ1Av1 =

1 11 10 0

(1/√

The remaining column vectors of U

must form an orthonormalbasis for N(AT ). We can compute a basis {x2, x3} for N(AT ) inthe usual way.

x2 = (1,−1, 0)T , x3 = (0, 0, 1)T

SVD. Examples

σ1Av1 =

1 11 10 0

(1/√

The remaining column vectors of U must form

an orthonormalbasis for N(AT ). We can compute a basis {x2, x3} for N(AT ) inthe usual way.

x2 = (1,−1, 0)T , x3 = (0, 0, 1)T

SVD. Examples

σ1Av1 =

1 11 10 0

(1/√

The remaining column vectors of U must form an orthonormal

basis for N(AT ). We can compute a basis {x2, x3} for N(AT ) inthe usual way.

x2 = (1,−1, 0)T , x3 = (0, 0, 1)T

SVD. Examples

σ1Av1 =

1 11 10 0

(1/√

The remaining column vectors of U must form an orthonormalbasis for

N(AT ). We can compute a basis {x2, x3} for N(AT ) inthe usual way.

x2 = (1,−1, 0)T , x3 = (0, 0, 1)T

SVD. Examples

σ1Av1 =

1 11 10 0

(1/√

The remaining column vectors of U must form an orthonormalbasis for N(AT ).

We can compute a basis {x2, x3} for N(AT ) inthe usual way.

x2 = (1,−1, 0)T , x3 = (0, 0, 1)T

SVD. Examples

σ1Av1 =

1 11 10 0

(1/√

The remaining column vectors of U must form an orthonormalbasis for N(AT ). We can compute

a basis {x2, x3} for N(AT ) inthe usual way.

x2 = (1,−1, 0)T , x3 = (0, 0, 1)T

SVD. Examples

σ1Av1 =

1 11 10 0

(1/√

The remaining column vectors of U must form an orthonormalbasis for N(AT ). We can compute a basis

{x2, x3} for N(AT ) inthe usual way.

x2 = (1,−1, 0)T , x3 = (0, 0, 1)T

SVD. Examples

σ1Av1 =

1 11 10 0

(1/√

The remaining column vectors of U must form an orthonormalbasis for N(AT ). We can compute a basis {x2, x3}

for N(AT ) inthe usual way.

x2 = (1,−1, 0)T , x3 = (0, 0, 1)T

SVD. Examples

σ1Av1 =

1 11 10 0

(1/√

The remaining column vectors of U must form an orthonormalbasis for N(AT ). We can compute a basis {x2, x3} for N(AT )

inthe usual way.

x2 = (1,−1, 0)T , x3 = (0, 0, 1)T

SVD. Examples

σ1Av1 =

1 11 10 0

(1/√

x2 = (1,−1, 0)T , x3 = (0, 0, 1)T

SVD. Examples

σ1Av1 =

1 11 10 0

(1/√

x2 = (1,−1, 0)T ,

x3 = (0, 0, 1)T

SVD. Examples

σ1Av1 =

1 11 10 0

(1/√

x2 = (1,−1, 0)T , x3 = (0, 0, 1)T

SVD. Examples

σ1Av1 =

1 11 10 0

(1/√

x2 = (1,−1, 0)T , x3 = (0, 0, 1)T

these vectors are already orthogonal, it is not necessary touse the Gram-Schmidt process to obtain an orthonormal basis.

SVD. Examples

σ1Av1 =

1 11 10 0

(1/√

x2 = (1,−1, 0)T , x3 = (0, 0, 1)T

Since these vectors

are already orthogonal, it is not necessary touse the Gram-Schmidt process to obtain an orthonormal basis.

SVD. Examples

σ1Av1 =

1 11 10 0

(1/√

x2 = (1,−1, 0)T , x3 = (0, 0, 1)T

Since these vectors are already orthogonal,

it is not necessary touse the Gram-Schmidt process to obtain an orthonormal basis.

SVD. Examples

σ1Av1 =

1 11 10 0

(1/√

x2 = (1,−1, 0)T , x3 = (0, 0, 1)T

Since these vectors are already orthogonal, it is not necessary

touse the Gram-Schmidt process to obtain an orthonormal basis.

SVD. Examples

σ1Av1 =

1 11 10 0

(1/√

x2 = (1,−1, 0)T , x3 = (0, 0, 1)T

Since these vectors are already orthogonal, it is not necessary touse

the Gram-Schmidt process to obtain an orthonormal basis.

SVD. Examples

σ1Av1 =

1 11 10 0

(1/√

x2 = (1,−1, 0)T , x3 = (0, 0, 1)T

Since these vectors are already orthogonal, it is not necessary touse the Gram-Schmidt process

to obtain an orthonormal basis.

SVD. Examples

σ1Av1 =

1 11 10 0

(1/√

x2 = (1,−1, 0)T , x3 = (0, 0, 1)T

Since these vectors are already orthogonal, it is not necessary touse the Gram-Schmidt process to obtain

an orthonormal basis.

SVD. Examples

σ1Av1 =

1 11 10 0

(1/√

x2 = (1,−1, 0)T , x3 = (0, 0, 1)T

SVD. Examples

We need only set

||x2||x2 = (

1√2,− 1√

2, 0)T , u3 =

||x3||x3 = (0, 0, 1)T

It then follows that

A = UΣV T =

01√2− 1√

0 00 0

(1√2

1√2− 1√

SVD. Examples

We need only

||x2||x2 = (

1√2,− 1√

2, 0)T , u3 =

||x3||x3 = (0, 0, 1)T

A = UΣV T =

01√2− 1√

0 00 0

(1√2

1√2− 1√

SVD. Examples

We need only set

||x2||x2 = (

1√2,− 1√

2, 0)T , u3 =

||x3||x3 = (0, 0, 1)T

A = UΣV T =

01√2− 1√

0 00 0

(1√2

1√2− 1√

SVD. Examples

We need only set

||x2||x2 = (

1√2,− 1√

2, 0)T , u3 =

||x3||x3 = (0, 0, 1)T

A = UΣV T =

01√2− 1√

0 00 0

(1√2

1√2− 1√

SVD. Examples

We need only set

||x2||x2 =

(1√2,− 1√

2, 0)T , u3 =

||x3||x3 = (0, 0, 1)T

A = UΣV T =

01√2− 1√

0 00 0

(1√2

1√2− 1√

SVD. Examples

We need only set

||x2||x2 = (

1√2,− 1√

2, 0)T , u3 =

||x3||x3 = (0, 0, 1)T

A = UΣV T =

01√2− 1√

0 00 0

(1√2

1√2− 1√

SVD. Examples

We need only set

||x2||x2 = (

1√2,− 1√

2, 0)T , u3 =

||x3||x3 =

(0, 0, 1)T

A = UΣV T =

01√2− 1√

0 00 0

(1√2

1√2− 1√

SVD. Examples

We need only set

||x2||x2 = (

1√2,− 1√

2, 0)T , u3 =

||x3||x3 = (0, 0, 1)T

A = UΣV T =

01√2− 1√

0 00 0

(1√2

1√2− 1√

SVD. Examples

We need only set

||x2||x2 = (

1√2,− 1√

2, 0)T , u3 =

||x3||x3 = (0, 0, 1)T

It then

follows that

A = UΣV T =

01√2− 1√

0 00 0

(1√2

1√2− 1√

SVD. Examples

We need only set

||x2||x2 = (

1√2,− 1√

2, 0)T , u3 =

||x3||x3 = (0, 0, 1)T

A = UΣV T =

01√2− 1√

0 00 0

(1√2

1√2− 1√

SVD. Examples

We need only set

||x2||x2 = (

1√2,− 1√

2, 0)T , u3 =

||x3||x3 = (0, 0, 1)T

UΣV T =

01√2− 1√

0 00 0

(1√2

1√2− 1√

SVD. Examples

We need only set

||x2||x2 = (

1√2,− 1√

2, 0)T , u3 =

||x3||x3 = (0, 0, 1)T

A = UΣV T =

01√2− 1√

0 00 0

(1√2

1√2− 1√

SVD. Examples

We need only set

||x2||x2 = (

1√2,− 1√

2, 0)T , u3 =

||x3||x3 = (0, 0, 1)T

A = UΣV T =

01√2− 1√

2 00 00 0

(1√2

1√2− 1√

SVD. Examples

We need only set

||x2||x2 = (

1√2,− 1√

2, 0)T , u3 =

||x3||x3 = (0, 0, 1)T

A = UΣV T =

01√2− 1√

0 00 0

(1√2

1√2− 1√

SVD. Examples

We need only set

||x2||x2 = (

1√2,− 1√

2, 0)T , u3 =

||x3||x3 = (0, 0, 1)T

A = UΣV T =

01√2− 1√

0 00 0

(1√2

1√2− 1√

SVD. Examples

We need only set

||x2||x2 = (

1√2,− 1√

2, 0)T , u3 =

||x3||x3 = (0, 0, 1)T

A = UΣV T =

01√2− 1√

0 00 0

(1√2

1√2− 1√

SVD. Examples

Example 12.8Let

(3 2 22 3 −2

Compute the singular values and the singular value decompositionof A.

Solution

The matrix

(17 88 17

SVD. Examples

Example 12.8

(3 2 22 3 −2

Solution

The matrix

(17 88 17

SVD. Examples

Example 12.8Let

(3 2 22 3 −2

Solution

The matrix

(17 88 17

SVD. Examples

Example 12.8Let

(3 2 22 3 −2

Solution

The matrix

(17 88 17

SVD. Examples

Example 12.8Let

(3 2 22 3 −2

Solution

The matrix

(17 88 17

SVD. Examples

Example 12.8Let

(3 2 22 3 −2

Compute

the singular values and the singular value decompositionof A.

Solution

The matrix

(17 88 17

SVD. Examples

Example 12.8Let

(3 2 22 3 −2

the singular value decompositionof A.

Solution

The matrix

(17 88 17

SVD. Examples

Example 12.8Let

(3 2 22 3 −2

Solution

The matrix

(17 88 17

SVD. Examples

Example 12.8Let

(3 2 22 3 −2

Solution

The matrix

(17 88 17

SVD. Examples

Example 12.8Let

(3 2 22 3 −2

Solution

The matrix

(17 88 17

SVD. Examples

Example 12.8Let

(3 2 22 3 −2

Solution

The matrix

(17 88 17

SVD. Examples

Example 12.8Let

(3 2 22 3 −2

Solution

The matrix

(17 88 17

SVD. Examples

Example 12.8Let

(3 2 22 3 −2

Solution

The matrix

(17 88 17

)Dr. Marco A Roque Sol Linear Algebra. Session 9

SVD. Examples

has eigenvalues λ1 = 25 and λ2 = 9. Consequently, the singularvalues of A are σ1 = 5 and σ2 = 3, and from here we can find theleft singular vectors (the columns of U ) but we’ll do it in adifferent way at the end.

Now we find the right singular vectors (the columns of V )byfinding an orthonormal set of eigenvectors of ATA. It is alsopossible to proceed by finding the left singular vectors (columns ofU) instead. The eigenvalues of ATA are 25, 9, and 0, and sinceATA is symmetric we know that the eigenvectors will beorthogonal.

For λ = 25, we have

ATA− 25I =

−12 12 212 −12 −22 −2 −17

SVD. Examples

has eigenvalues

λ1 = 25 and λ2 = 9. Consequently, the singularvalues of A are σ1 = 5 and σ2 = 3, and from here we can find theleft singular vectors (the columns of U ) but we’ll do it in adifferent way at the end.

ATA− 25I =

−12 12 212 −12 −22 −2 −17

SVD. Examples

has eigenvalues λ1 = 25 and λ2 = 9.

Consequently, the singularvalues of A are σ1 = 5 and σ2 = 3, and from here we can find theleft singular vectors (the columns of U ) but we’ll do it in adifferent way at the end.

ATA− 25I =

−12 12 212 −12 −22 −2 −17

SVD. Examples

has eigenvalues λ1 = 25 and λ2 = 9. Consequently,

the singularvalues of A are σ1 = 5 and σ2 = 3, and from here we can find theleft singular vectors (the columns of U ) but we’ll do it in adifferent way at the end.

ATA− 25I =

−12 12 212 −12 −22 −2 −17

SVD. Examples

has eigenvalues λ1 = 25 and λ2 = 9. Consequently, the singularvalues

of A are σ1 = 5 and σ2 = 3, and from here we can find theleft singular vectors (the columns of U ) but we’ll do it in adifferent way at the end.

ATA− 25I =

−12 12 212 −12 −22 −2 −17

SVD. Examples

has eigenvalues λ1 = 25 and λ2 = 9. Consequently, the singularvalues of A are

σ1 = 5 and σ2 = 3, and from here we can find theleft singular vectors (the columns of U ) but we’ll do it in adifferent way at the end.

ATA− 25I =

−12 12 212 −12 −22 −2 −17

SVD. Examples

has eigenvalues λ1 = 25 and λ2 = 9. Consequently, the singularvalues of A are σ1 = 5 and

σ2 = 3, and from here we can find theleft singular vectors (the columns of U ) but we’ll do it in adifferent way at the end.

ATA− 25I =

−12 12 212 −12 −22 −2 −17

SVD. Examples

has eigenvalues λ1 = 25 and λ2 = 9. Consequently, the singularvalues of A are σ1 = 5 and

σ2 = 3, and from here we can find theleft singular vectors (the columns of U ) but we’ll do it in adifferent way at the end.

ATA− 25I =

−12 12 212 −12 −22 −2 −17

SVD. Examples

has eigenvalues λ1 = 25 and λ2 = 9. Consequently, the singularvalues of A are σ1 = 5 and σ2 = 3, and from here we can find theleft singular vectors

(the columns of U ) but we’ll do it in adifferent way at the end.

ATA− 25I =

−12 12 212 −12 −22 −2 −17

SVD. Examples

ATA− 25I =

−12 12 212 −12 −22 −2 −17

SVD. Examples

we find the right singular vectors (the columns of V )byfinding an orthonormal set of eigenvectors of ATA. It is alsopossible to proceed by finding the left singular vectors (columns ofU) instead. The eigenvalues of ATA are 25, 9, and 0, and sinceATA is symmetric we know that the eigenvectors will beorthogonal.

ATA− 25I =

−12 12 212 −12 −22 −2 −17

SVD. Examples

Now we find

the right singular vectors (the columns of V )byfinding an orthonormal set of eigenvectors of ATA. It is alsopossible to proceed by finding the left singular vectors (columns ofU) instead. The eigenvalues of ATA are 25, 9, and 0, and sinceATA is symmetric we know that the eigenvectors will beorthogonal.

ATA− 25I =

−12 12 212 −12 −22 −2 −17

SVD. Examples

Now we find the right singular vectors

(the columns of V )byfinding an orthonormal set of eigenvectors of ATA. It is alsopossible to proceed by finding the left singular vectors (columns ofU) instead. The eigenvalues of ATA are 25, 9, and 0, and sinceATA is symmetric we know that the eigenvectors will beorthogonal.

ATA− 25I =

−12 12 212 −12 −22 −2 −17

SVD. Examples

Now we find the right singular vectors (the columns of V )

byfinding an orthonormal set of eigenvectors of ATA. It is alsopossible to proceed by finding the left singular vectors (columns ofU) instead. The eigenvalues of ATA are 25, 9, and 0, and sinceATA is symmetric we know that the eigenvectors will beorthogonal.

ATA− 25I =

−12 12 212 −12 −22 −2 −17

SVD. Examples

Now we find the right singular vectors (the columns of V )byfinding

an orthonormal set of eigenvectors of ATA. It is alsopossible to proceed by finding the left singular vectors (columns ofU) instead. The eigenvalues of ATA are 25, 9, and 0, and sinceATA is symmetric we know that the eigenvectors will beorthogonal.

ATA− 25I =

−12 12 212 −12 −22 −2 −17

SVD. Examples

Now we find the right singular vectors (the columns of V )byfinding an orthonormal set of eigenvectors

of ATA. It is alsopossible to proceed by finding the left singular vectors (columns ofU) instead. The eigenvalues of ATA are 25, 9, and 0, and sinceATA is symmetric we know that the eigenvectors will beorthogonal.

ATA− 25I =

−12 12 212 −12 −22 −2 −17

SVD. Examples

Now we find the right singular vectors (the columns of V )byfinding an orthonormal set of eigenvectors of ATA.

It is alsopossible to proceed by finding the left singular vectors (columns ofU) instead. The eigenvalues of ATA are 25, 9, and 0, and sinceATA is symmetric we know that the eigenvectors will beorthogonal.

ATA− 25I =

−12 12 212 −12 −22 −2 −17

SVD. Examples

Now we find the right singular vectors (the columns of V )byfinding an orthonormal set of eigenvectors of ATA. It is alsopossible

to proceed by finding the left singular vectors (columns ofU) instead. The eigenvalues of ATA are 25, 9, and 0, and sinceATA is symmetric we know that the eigenvectors will beorthogonal.

ATA− 25I =

−12 12 212 −12 −22 −2 −17

SVD. Examples

Now we find the right singular vectors (the columns of V )byfinding an orthonormal set of eigenvectors of ATA. It is alsopossible to proceed by finding

the left singular vectors (columns ofU) instead. The eigenvalues of ATA are 25, 9, and 0, and sinceATA is symmetric we know that the eigenvectors will beorthogonal.

ATA− 25I =

−12 12 212 −12 −22 −2 −17

SVD. Examples

Now we find the right singular vectors (the columns of V )byfinding an orthonormal set of eigenvectors of ATA. It is alsopossible to proceed by finding the left singular vectors

(columns ofU) instead. The eigenvalues of ATA are 25, 9, and 0, and sinceATA is symmetric we know that the eigenvectors will beorthogonal.

ATA− 25I =

−12 12 212 −12 −22 −2 −17

SVD. Examples

Now we find the right singular vectors (the columns of V )byfinding an orthonormal set of eigenvectors of ATA. It is alsopossible to proceed by finding the left singular vectors (columns ofU)

instead. The eigenvalues of ATA are 25, 9, and 0, and sinceATA is symmetric we know that the eigenvectors will beorthogonal.

ATA− 25I =

−12 12 212 −12 −22 −2 −17

SVD. Examples

Now we find the right singular vectors (the columns of V )byfinding an orthonormal set of eigenvectors of ATA. It is alsopossible to proceed by finding the left singular vectors (columns ofU) instead.

The eigenvalues of ATA are 25, 9, and 0, and sinceATA is symmetric we know that the eigenvectors will beorthogonal.

ATA− 25I =

−12 12 212 −12 −22 −2 −17

SVD. Examples

Now we find the right singular vectors (the columns of V )byfinding an orthonormal set of eigenvectors of ATA. It is alsopossible to proceed by finding the left singular vectors (columns ofU) instead. The eigenvalues

of ATA are 25, 9, and 0, and sinceATA is symmetric we know that the eigenvectors will beorthogonal.

ATA− 25I =

−12 12 212 −12 −22 −2 −17

SVD. Examples

Now we find the right singular vectors (the columns of V )byfinding an orthonormal set of eigenvectors of ATA. It is alsopossible to proceed by finding the left singular vectors (columns ofU) instead. The eigenvalues of ATA

are 25, 9, and 0, and sinceATA is symmetric we know that the eigenvectors will beorthogonal.

ATA− 25I =

−12 12 212 −12 −22 −2 −17

SVD. Examples

Now we find the right singular vectors (the columns of V )byfinding an orthonormal set of eigenvectors of ATA. It is alsopossible to proceed by finding the left singular vectors (columns ofU) instead. The eigenvalues of ATA are 25, 9, and 0, and

sinceATA is symmetric we know that the eigenvectors will beorthogonal.

ATA− 25I =

−12 12 212 −12 −22 −2 −17

SVD. Examples

Now we find the right singular vectors (the columns of V )byfinding an orthonormal set of eigenvectors of ATA. It is alsopossible to proceed by finding the left singular vectors (columns ofU) instead. The eigenvalues of ATA are 25, 9, and 0, and sinceATA is symmetric

we know that the eigenvectors will beorthogonal.

ATA− 25I =

−12 12 212 −12 −22 −2 −17

SVD. Examples

Now we find the right singular vectors (the columns of V )byfinding an orthonormal set of eigenvectors of ATA. It is alsopossible to proceed by finding the left singular vectors (columns ofU) instead. The eigenvalues of ATA are 25, 9, and 0, and sinceATA is symmetric we know that

the eigenvectors will beorthogonal.

ATA− 25I =

−12 12 212 −12 −22 −2 −17

SVD. Examples

ATA− 25I =

−12 12 212 −12 −22 −2 −17

SVD. Examples

For λ = 25,

we have

ATA− 25I =

−12 12 212 −12 −22 −2 −17

SVD. Examples

ATA− 25I =

−12 12 212 −12 −22 −2 −17

SVD. Examples

ATA− 25I =

−12 12 212 −12 −22 −2 −17

SVD. Examples

ATA− 25I =

−12 12 212 −12 −22 −2 −17

SVD. Examples

which row-reduces to 1 −1 00 0 10 0 0

A unit-length vector in the kernel of that matrix

v1 =1√2

For λ = 9, we have

ATA− 9I =

4 12 212 4 −22 −2 −1

SVD. Examples

v1 =1√2

For λ = 9, we have

ATA− 9I =

4 12 212 4 −22 −2 −1

SVD. Examples

A unit-length vector

in the kernel of that matrix

v1 =1√2

For λ = 9, we have

ATA− 9I =

4 12 212 4 −22 −2 −1

SVD. Examples

v1 =1√2

For λ = 9, we have

ATA− 9I =

4 12 212 4 −22 −2 −1

SVD. Examples

For λ = 9, we have

ATA− 9I =

4 12 212 4 −22 −2 −1

SVD. Examples

v1 =1√2

For λ = 9, we have

ATA− 9I =

4 12 212 4 −22 −2 −1

SVD. Examples

v1 =1√2

For λ = 9, we have

ATA− 9I =

4 12 212 4 −22 −2 −1

SVD. Examples

v1 =1√2

For λ = 9,

we have

ATA− 9I =

4 12 212 4 −22 −2 −1

SVD. Examples

v1 =1√2

For λ = 9, we have

ATA− 9I =

4 12 212 4 −22 −2 −1

SVD. Examples

v1 =1√2

For λ = 9, we have

ATA− 9I =

4 12 212 4 −22 −2 −1

SVD. Examples

v1 =1√2

For λ = 9, we have

ATA− 9I =

4 12 212 4 −22 −2 −1

SVD. Examples

which row-reduces to 1 0 −1/40 1 1/40 0 0

v2 =1√18

1−14

For the last eigenvector, we could compute the kernel of ATA orfind a unit vector perpendicular to v1 and v2. To be perpendicularto v1, vT3 = (a, b, c), must satisfy that −a = b and to beperpendicular to v2, vT3 = (a,−a, c), must satisfy that2√18

+ 4c√18

SVD. Examples

v2 =1√18

1−14

+ 4c√18

SVD. Examples

A unit-length vector

in the kernel of that matrix

v2 =1√18

1−14

+ 4c√18

SVD. Examples

v2 =1√18

1−14

+ 4c√18

SVD. Examples

1√18

1−14

+ 4c√18

SVD. Examples

v2 =1√18

1−14

+ 4c√18

SVD. Examples

v2 =1√18

1−14

+ 4c√18

SVD. Examples

v2 =1√18

1−14

For the last eigenvector,

we could compute the kernel of ATA orfind a unit vector perpendicular to v1 and v2. To be perpendicularto v1, vT3 = (a, b, c), must satisfy that −a = b and to beperpendicular to v2, vT3 = (a,−a, c), must satisfy that2√18

+ 4c√18

SVD. Examples

v2 =1√18

1−14

For the last eigenvector, we could compute

the kernel of ATA orfind a unit vector perpendicular to v1 and v2. To be perpendicularto v1, vT3 = (a, b, c), must satisfy that −a = b and to beperpendicular to v2, vT3 = (a,−a, c), must satisfy that2√18

+ 4c√18

SVD. Examples

v2 =1√18

1−14

For the last eigenvector, we could compute the kernel of ATA or

find a unit vector perpendicular to v1 and v2. To be perpendicularto v1, vT3 = (a, b, c), must satisfy that −a = b and to beperpendicular to v2, vT3 = (a,−a, c), must satisfy that2√18

+ 4c√18

SVD. Examples

v2 =1√18

1−14

For the last eigenvector, we could compute the kernel of ATA orfind a unit vector

perpendicular to v1 and v2. To be perpendicularto v1, vT3 = (a, b, c), must satisfy that −a = b and to beperpendicular to v2, vT3 = (a,−a, c), must satisfy that2√18

+ 4c√18

SVD. Examples

v2 =1√18

1−14

For the last eigenvector, we could compute the kernel of ATA orfind a unit vector perpendicular to v1 and v2.

To be perpendicularto v1, vT3 = (a, b, c), must satisfy that −a = b and to beperpendicular to v2, vT3 = (a,−a, c), must satisfy that2√18

+ 4c√18

SVD. Examples

v2 =1√18

1−14

For the last eigenvector, we could compute the kernel of ATA orfind a unit vector perpendicular to v1 and v2. To be perpendicular

to v1, vT3 = (a, b, c), must satisfy that −a = b and to beperpendicular to v2, vT3 = (a,−a, c), must satisfy that2√18

+ 4c√18

SVD. Examples

v2 =1√18

1−14

For the last eigenvector, we could compute the kernel of ATA orfind a unit vector perpendicular to v1 and v2. To be perpendicularto v1, vT3 = (a, b, c),

must satisfy that −a = b and to beperpendicular to v2, vT3 = (a,−a, c), must satisfy that2√18

+ 4c√18

SVD. Examples

v2 =1√18

1−14

For the last eigenvector, we could compute the kernel of ATA orfind a unit vector perpendicular to v1 and v2. To be perpendicularto v1, vT3 = (a, b, c), must satisfy that

−a = b and to beperpendicular to v2, vT3 = (a,−a, c), must satisfy that2√18

+ 4c√18

SVD. Examples

v2 =1√18

1−14

For the last eigenvector, we could compute the kernel of ATA orfind a unit vector perpendicular to v1 and v2. To be perpendicularto v1, vT3 = (a, b, c), must satisfy that −a = b and

to beperpendicular to v2, vT3 = (a,−a, c), must satisfy that2√18

+ 4c√18

SVD. Examples

v2 =1√18

1−14

For the last eigenvector, we could compute the kernel of ATA orfind a unit vector perpendicular to v1 and v2. To be perpendicularto v1, vT3 = (a, b, c), must satisfy that −a = b and to beperpendicular

to v2, vT3 = (a,−a, c), must satisfy that2√18

+ 4c√18

SVD. Examples

v2 =1√18

1−14

For the last eigenvector, we could compute the kernel of ATA orfind a unit vector perpendicular to v1 and v2. To be perpendicularto v1, vT3 = (a, b, c), must satisfy that −a = b and to beperpendicular to v2,

vT3 = (a,−a, c), must satisfy that2√18

+ 4c√18

SVD. Examples

v2 =1√18

1−14

For the last eigenvector, we could compute the kernel of ATA orfind a unit vector perpendicular to v1 and v2. To be perpendicularto v1, vT3 = (a, b, c), must satisfy that −a = b and to beperpendicular to v2, vT3 = (a,−a, c),

must satisfy that2√18

+ 4c√18

SVD. Examples

v2 =1√18

1−14

For the last eigenvector, we could compute the kernel of ATA orfind a unit vector perpendicular to v1 and v2. To be perpendicularto v1, vT3 = (a, b, c), must satisfy that −a = b and to beperpendicular to v2, vT3 = (a,−a, c), must satisfy that

2√18

+ 4c√18

SVD. Examples

v2 =1√18

1−14

+ 4c√18

= 0.Dr. Marco A Roque Sol Linear Algebra. Session 9

SVD. Examples

v2 =1√18

a−a−a/2

and for it to be unit-length we need a = 2/3

v3 =1√18

2/3−2/3−1/3

So at this point we know that

A = UΣV T = U

(5 0 00 3 0

) 1/√

2 1/√

18 −1/√

19 4/√

182/3 −2/3 −1/3

SVD. Examples

1√18

a−a−a/2

v3 =1√18

2/3−2/3−1/3

A = UΣV T = U

(5 0 00 3 0

) 1/√

2 1/√

18 −1/√

19 4/√

182/3 −2/3 −1/3

SVD. Examples

v2 =1√18

a−a−a/2

v3 =1√18

2/3−2/3−1/3

A = UΣV T = U

(5 0 00 3 0

) 1/√

2 1/√

18 −1/√

19 4/√

182/3 −2/3 −1/3

SVD. Examples

v2 =1√18

a−a−a/2

v3 =1√18

2/3−2/3−1/3

A = UΣV T = U

(5 0 00 3 0

) 1/√

2 1/√

18 −1/√

19 4/√

182/3 −2/3 −1/3

SVD. Examples

v2 =1√18

a−a−a/2

and for it

to be unit-length we need a = 2/3

v3 =1√18

2/3−2/3−1/3

A = UΣV T = U

(5 0 00 3 0

) 1/√

2 1/√

18 −1/√

19 4/√

182/3 −2/3 −1/3

SVD. Examples

v2 =1√18

a−a−a/2

and for it to be unit-length

we need a = 2/3

v3 =1√18

2/3−2/3−1/3

A = UΣV T = U

(5 0 00 3 0

) 1/√

2 1/√

18 −1/√

19 4/√

182/3 −2/3 −1/3

SVD. Examples

v2 =1√18

a−a−a/2

v3 =1√18

2/3−2/3−1/3

A = UΣV T = U

(5 0 00 3 0

) 1/√

2 1/√

18 −1/√

19 4/√

182/3 −2/3 −1/3

SVD. Examples

v2 =1√18

a−a−a/2

1√18

2/3−2/3−1/3

A = UΣV T = U

(5 0 00 3 0

) 1/√

2 1/√

18 −1/√

19 4/√

182/3 −2/3 −1/3

SVD. Examples

v2 =1√18

a−a−a/2

v3 =1√18

2/3−2/3−1/3

A = UΣV T = U

(5 0 00 3 0

) 1/√

2 1/√

18 −1/√

19 4/√

182/3 −2/3 −1/3

SVD. Examples

v2 =1√18

a−a−a/2

v3 =1√18

2/3−2/3−1/3

A = UΣV T = U

(5 0 00 3 0

) 1/√

2 1/√

18 −1/√

19 4/√

182/3 −2/3 −1/3

SVD. Examples

v2 =1√18

a−a−a/2

v3 =1√18

2/3−2/3−1/3

So at this point

we know that

A = UΣV T = U

(5 0 00 3 0

) 1/√

2 1/√

18 −1/√

19 4/√

182/3 −2/3 −1/3

SVD. Examples

v2 =1√18

a−a−a/2

v3 =1√18

2/3−2/3−1/3

A = UΣV T = U

(5 0 00 3 0

) 1/√

2 1/√

18 −1/√

19 4/√

182/3 −2/3 −1/3

SVD. Examples

v2 =1√18

a−a−a/2

v3 =1√18

2/3−2/3−1/3

A = UΣV T = U

(5 0 00 3 0

) 1/√

2 1/√

18 −1/√

19 4/√

182/3 −2/3 −1/3

SVD. Examples

Finally, we can compute U by the formula σui = Avi orui = 1

σAvi . This gives

(1/√

2 1/√

2 −1/√

)So the full SVD is:

A = UΣV T =(1/√

2 1/√

2 −1/√

)(5 0 00 3 0

) 1/√

2 1/√

18 −1/√

19 4/√

182/3 −2/3 −1/3

SVD. Examples

Finally,

we can compute U by the formula σui = Avi orui = 1

σAvi . This gives

(1/√

2 1/√

2 −1/√

A = UΣV T =(1/√

2 1/√

2 −1/√

)(5 0 00 3 0

) 1/√

2 1/√

18 −1/√

19 4/√

182/3 −2/3 −1/3

SVD. Examples

Finally, we can compute U

by the formula σui = Avi orui = 1

σAvi . This gives

(1/√

2 1/√

2 −1/√

A = UΣV T =(1/√

2 1/√

2 −1/√

)(5 0 00 3 0

) 1/√

2 1/√

18 −1/√

19 4/√

182/3 −2/3 −1/3

SVD. Examples

Finally, we can compute U by the formula σui = Avi or

ui = 1σAvi . This gives

(1/√

2 1/√

2 −1/√

A = UΣV T =(1/√

2 1/√

2 −1/√

)(5 0 00 3 0

) 1/√

2 1/√

18 −1/√

19 4/√

182/3 −2/3 −1/3

SVD. Examples

σAvi .

This gives

(1/√

2 1/√

2 −1/√

A = UΣV T =(1/√

2 1/√

2 −1/√

)(5 0 00 3 0

) 1/√

2 1/√

18 −1/√

19 4/√

182/3 −2/3 −1/3

SVD. Examples

σAvi . This gives

(1/√

2 1/√

2 −1/√

A = UΣV T =(1/√

2 1/√

2 −1/√

)(5 0 00 3 0

) 1/√

2 1/√

18 −1/√

19 4/√

182/3 −2/3 −1/3

SVD. Examples

σAvi . This gives

(1/√

2 1/√

2 −1/√

So the full SVD is:

A = UΣV T =(1/√

2 1/√

2 −1/√

)(5 0 00 3 0

) 1/√

2 1/√

18 −1/√

19 4/√

182/3 −2/3 −1/3

SVD. Examples

σAvi . This gives

(1/√

2 1/√

2 −1/√

the full SVD is:

A = UΣV T =(1/√

2 1/√

2 −1/√

)(5 0 00 3 0

) 1/√

2 1/√

18 −1/√

19 4/√

182/3 −2/3 −1/3

SVD. Examples

σAvi . This gives

(1/√

2 1/√

2 −1/√

A = UΣV T =(1/√

2 1/√

2 −1/√

)(5 0 00 3 0

) 1/√

2 1/√

18 −1/√

19 4/√

182/3 −2/3 −1/3

SVD. Examples

σAvi . This gives

(1/√

2 1/√

2 −1/√

A = UΣV T =

(1/√

2 1/√

2 −1/√

)(5 0 00 3 0

) 1/√

2 1/√

18 −1/√

19 4/√

182/3 −2/3 −1/3

SVD. Examples

σAvi . This gives

(1/√

2 1/√

2 −1/√

A = UΣV T =(1/√

2 1/√

2 −1/√

)(5 0 00 3 0

) 1/√

2 1/√

18 −1/√

19 4/√

182/3 −2/3 −1/3

SVD. Examples

If A has singular value decomposition UΣV T , then A can berepresented by the outer product expansion

A = σ1u1vT1 + σ2u2vT2 + · · ·+ σnunvTn

The closest matrix of rank k , is obtained by truncating this sum,after the first k terms:

A′ = σ1u1vT1 + σ1u2vT2 + · · ·+ σkukvTk , k < n

SVD. Examples

has singular value decomposition UΣV T , then A can berepresented by the outer product expansion

SVD. Examples

If A has singular value decomposition

UΣV T , then A can berepresented by the outer product expansion

SVD. Examples

If A has singular value decomposition UΣV T ,

then A can berepresented by the outer product expansion

SVD. Examples

If A has singular value decomposition UΣV T , then A

can berepresented by the outer product expansion

SVD. Examples

If A has singular value decomposition UΣV T , then A can berepresented

by the outer product expansion

SVD. Examples

If A has singular value decomposition UΣV T , then A can berepresented by the

outer product expansion

SVD. Examples

The closest matrix

of rank k , is obtained by truncating this sum,after the first k terms:

SVD. Examples

The closest matrix of rank k ,

is obtained by truncating this sum,after the first k terms:

SVD. Examples

The closest matrix of rank k , is obtained by

truncating this sum,after the first k terms:

SVD. Examples

The closest matrix of rank k , is obtained by truncating this sum,

after the first k terms:

SVD. Examples

The closest matrix of rank k , is obtained by truncating this sum,after the first

k terms:

SVD. Examples

Example 12.9Let

0 1 1√2 2 0

Solution

The matrix

2 2 22 6 22 2 2

SVD. Examples

Example 12.9

0 1 1√2 2 0

Solution

The matrix

2 2 22 6 22 2 2

SVD. Examples

Example 12.9Let

0 1 1√2 2 0

Solution

The matrix

2 2 22 6 22 2 2

SVD. Examples

Example 12.9Let

0 1 1√2 2 0

Solution

The matrix

2 2 22 6 22 2 2

SVD. Examples

Example 12.9Let

0 1 1√2 2 0

Solution

The matrix

2 2 22 6 22 2 2

SVD. Examples

Example 12.9Let

0 1 1√2 2 0

Compute

the singular values and the singular value decompositionof A.

Solution

The matrix

2 2 22 6 22 2 2

SVD. Examples

Example 12.9Let

0 1 1√2 2 0

the singular value decompositionof A.

Solution

The matrix

2 2 22 6 22 2 2

SVD. Examples

Example 12.9Let

0 1 1√2 2 0

Solution

The matrix

2 2 22 6 22 2 2

SVD. Examples

Example 12.9Let

0 1 1√2 2 0

Solution

The matrix

2 2 22 6 22 2 2

SVD. Examples

Example 12.9Let

0 1 1√2 2 0

Solution

The matrix

2 2 22 6 22 2 2

SVD. Examples

Example 12.9Let

0 1 1√2 2 0

Solution

The matrix

2 2 22 6 22 2 2

SVD. Examples

Example 12.9Let

0 1 1√2 2 0

Solution

The matrix

2 2 22 6 22 2 2

SVD. Examples

Example 12.9Let

0 1 1√2 2 0

Solution

The matrix

2 2 22 6 22 2 2

SVD. Examples

The characteristic polynomial is

−λ3 + 10λ2− 16λ = −λ(λ2− 10λ+ 16) = −λ(λ− 8)(λ− 2) = 0

and AAT has eigenvalues λ1 = 8, λ2 = 2 and λ3 = 0.Consequently, the singular values of A are σ1 = 2

√2, σ2 =

√2 and

σ3 = 0

To give the decomposition, we consider the diagonal matrix ofsingular values

2 00 0 0

SVD. Examples

−λ3 + 10λ2− 16λ = −λ(λ2− 10λ+ 16) = −λ(λ− 8)(λ− 2) = 0

√2, σ2 =

√2 and

σ3 = 0

2 00 0 0

SVD. Examples

−λ3 + 10λ2− 16λ =

−λ(λ2− 10λ+ 16) = −λ(λ− 8)(λ− 2) = 0

√2, σ2 =

√2 and

σ3 = 0

2 00 0 0

SVD. Examples

−λ3 + 10λ2− 16λ = −λ(λ2− 10λ+ 16) =

−λ(λ− 8)(λ− 2) = 0

√2, σ2 =

√2 and

σ3 = 0

2 00 0 0

SVD. Examples

−λ3 + 10λ2− 16λ = −λ(λ2− 10λ+ 16) = −λ(λ− 8)(λ− 2) = 0

√2, σ2 =

√2 and

σ3 = 0

2 00 0 0

SVD. Examples

−λ3 + 10λ2− 16λ = −λ(λ2− 10λ+ 16) = −λ(λ− 8)(λ− 2) = 0

and AAT has eigenvalues

λ1 = 8, λ2 = 2 and λ3 = 0.Consequently, the singular values of A are σ1 = 2

√2, σ2 =

√2 and

σ3 = 0

2 00 0 0

SVD. Examples

−λ3 + 10λ2− 16λ = −λ(λ2− 10λ+ 16) = −λ(λ− 8)(λ− 2) = 0

and AAT has eigenvalues λ1 = 8,

λ2 = 2 and λ3 = 0.Consequently, the singular values of A are σ1 = 2

√2, σ2 =

√2 and

σ3 = 0

2 00 0 0

SVD. Examples

−λ3 + 10λ2− 16λ = −λ(λ2− 10λ+ 16) = −λ(λ− 8)(λ− 2) = 0

and AAT has eigenvalues λ1 = 8, λ2 = 2 and

λ3 = 0.Consequently, the singular values of A are σ1 = 2

√2, σ2 =

√2 and

σ3 = 0

2 00 0 0

SVD. Examples

−λ3 + 10λ2− 16λ = −λ(λ2− 10λ+ 16) = −λ(λ− 8)(λ− 2) = 0

and AAT has eigenvalues λ1 = 8, λ2 = 2 and λ3 = 0.

Consequently, the singular values of A are σ1 = 2√

2, σ2 =√

2 andσ3 = 0

2 00 0 0

SVD. Examples

−λ3 + 10λ2− 16λ = −λ(λ2− 10λ+ 16) = −λ(λ− 8)(λ− 2) = 0

and AAT has eigenvalues λ1 = 8, λ2 = 2 and λ3 = 0.Consequently,

the singular values of A are σ1 = 2√

2, σ2 =√

2 andσ3 = 0

2 00 0 0

SVD. Examples

−λ3 + 10λ2− 16λ = −λ(λ2− 10λ+ 16) = −λ(λ− 8)(λ− 2) = 0

and AAT has eigenvalues λ1 = 8, λ2 = 2 and λ3 = 0.Consequently, the singular values

of A are σ1 = 2√

2, σ2 =√

2 andσ3 = 0

2 00 0 0

SVD. Examples

−λ3 + 10λ2− 16λ = −λ(λ2− 10λ+ 16) = −λ(λ− 8)(λ− 2) = 0

and AAT has eigenvalues λ1 = 8, λ2 = 2 and λ3 = 0.Consequently, the singular values of A are

σ1 = 2√

2, σ2 =√

2 andσ3 = 0

2 00 0 0

SVD. Examples

−λ3 + 10λ2− 16λ = −λ(λ2− 10λ+ 16) = −λ(λ− 8)(λ− 2) = 0

σ2 =√

2 andσ3 = 0

2 00 0 0

SVD. Examples

−λ3 + 10λ2− 16λ = −λ(λ2− 10λ+ 16) = −λ(λ− 8)(λ− 2) = 0

√2, σ2 =

√2 and

σ3 = 0

2 00 0 0

SVD. Examples

−λ3 + 10λ2− 16λ = −λ(λ2− 10λ+ 16) = −λ(λ− 8)(λ− 2) = 0

√2, σ2 =

√2 and

σ3 = 0

2 00 0 0

SVD. Examples

−λ3 + 10λ2− 16λ = −λ(λ2− 10λ+ 16) = −λ(λ− 8)(λ− 2) = 0

√2, σ2 =

√2 and

σ3 = 0

2 00 0 0

SVD. Examples

−λ3 + 10λ2− 16λ = −λ(λ2− 10λ+ 16) = −λ(λ− 8)(λ− 2) = 0

√2, σ2 =

√2 and

σ3 = 0

2 00 0 0

SVD. Examples

−λ3 + 10λ2− 16λ = −λ(λ2− 10λ+ 16) = −λ(λ− 8)(λ− 2) = 0

√2, σ2 =

√2 and

σ3 = 0

2 00 0 0

SVD. Examples

Next, we find an orthonormal set of eigenvectors for AAT . Theeigenvalues of AAT are 8, 2, and 0, and since AAT is symmetric weknow that the eigenvectors will be orthogonal. This giveseigenvectors

1√62√61√6

; u2 =

−1√3

− 1√3

0− 1√

SVD. Examples

Next, we find an orthonormal set of eigenvectors for AAT . Theeigenvalues

of AAT are 8, 2, and 0, and since AAT is symmetric weknow that the eigenvectors will be orthogonal. This giveseigenvectors

1√62√61√6

; u2 =

−1√3

− 1√3

0− 1√

SVD. Examples

Next, we find an orthonormal set of eigenvectors for AAT . Theeigenvalues of AAT

are 8, 2, and 0, and since AAT is symmetric weknow that the eigenvectors will be orthogonal. This giveseigenvectors

1√62√61√6

; u2 =

−1√3

− 1√3

0− 1√

SVD. Examples

Next, we find an orthonormal set of eigenvectors for AAT . Theeigenvalues of AAT are 8, 2, and 0, and

since AAT is symmetric weknow that the eigenvectors will be orthogonal. This giveseigenvectors

1√62√61√6

; u2 =

−1√3

− 1√3

0− 1√

SVD. Examples

Next, we find an orthonormal set of eigenvectors for AAT . Theeigenvalues of AAT are 8, 2, and 0, and since AAT is symmetric

weknow that the eigenvectors will be orthogonal. This giveseigenvectors

1√62√61√6

; u2 =

−1√3

− 1√3

0− 1√

SVD. Examples

Next, we find an orthonormal set of eigenvectors for AAT . Theeigenvalues of AAT are 8, 2, and 0, and since AAT is symmetric weknow that

the eigenvectors will be orthogonal. This giveseigenvectors

1√62√61√6

; u2 =

−1√3

− 1√3

0− 1√

SVD. Examples

Next, we find an orthonormal set of eigenvectors for AAT . Theeigenvalues of AAT are 8, 2, and 0, and since AAT is symmetric weknow that the eigenvectors will be orthogonal.

This giveseigenvectors

1√62√61√6

; u2 =

−1√3

− 1√3

0− 1√

SVD. Examples

1√62√61√6

; u2 =

−1√3

− 1√3

0− 1√

SVD. Examples

1√62√61√6

; u2 =

−1√3

− 1√3

0− 1√

SVD. Examples

1√62√61√6

−1√3

− 1√3

0− 1√

SVD. Examples

1√62√61√6

; u2 =

−1√3

− 1√3

0− 1√

SVD. Examples

1√62√61√6

; u2 =

−1√3

− 1√3

0− 1√

SVD. Examples

1√62√61√6

; u2 =

−1√3

− 1√3

0− 1√

SVD. Examples

1√62√61√6

; u2 =

−1√3

− 1√3

0− 1√

SVD. Examples

Put these together to get

1√6− 1√

31√2

01√6− 1√

3− 1√

2 2√

2 6 20 2 2

The eigenvalues of ATA are 8, 2, and 0, and since ATA issymmetric we know that the eigenvectors will be orthogonal. Thisgives eigenvectors

SVD. Examples

1√6− 1√

31√2

01√6− 1√

3− 1√

2 2√

2 6 20 2 2

SVD. Examples

1√6− 1√

31√2

01√6− 1√

3− 1√

2 2√

2 6 20 2 2

SVD. Examples

1√6− 1√

31√2

01√6− 1√

3− 1√

2 2√

2 6 20 2 2

SVD. Examples

1√6− 1√

31√2

01√6− 1√

3− 1√

we find the right singular vectors (the columns of V )byfinding an orthonormal set of eigenvectors of ATA.

2 2√

2 6 20 2 2

SVD. Examples

1√6− 1√

31√2

01√6− 1√

3− 1√

Now we find

the right singular vectors (the columns of V )byfinding an orthonormal set of eigenvectors of ATA.

2 2√

2 6 20 2 2

SVD. Examples

1√6− 1√

31√2

01√6− 1√

3− 1√

Now we find the right singular vectors

(the columns of V )byfinding an orthonormal set of eigenvectors of ATA.

2 2√

2 6 20 2 2

SVD. Examples

1√6− 1√

31√2

01√6− 1√

3− 1√

Now we find the right singular vectors (the columns of V )

byfinding an orthonormal set of eigenvectors of ATA.

2 2√

2 6 20 2 2

SVD. Examples

1√6− 1√

31√2

01√6− 1√

3− 1√

Now we find the right singular vectors (the columns of V )byfinding

an orthonormal set of eigenvectors of ATA.

2 2√

2 6 20 2 2

SVD. Examples

1√6− 1√

31√2

01√6− 1√

3− 1√

Now we find the right singular vectors (the columns of V )byfinding an orthonormal set of eigenvectors

of ATA.

2 2√

2 6 20 2 2

SVD. Examples

1√6− 1√

31√2

01√6− 1√

3− 1√

2 2√

2 6 20 2 2

SVD. Examples

1√6− 1√

31√2

01√6− 1√

3− 1√

2 2√

2 6 20 2 2

SVD. Examples

1√6− 1√

31√2

01√6− 1√

3− 1√

2 2√

2 6 20 2 2

SVD. Examples

1√6− 1√

31√2

01√6− 1√

3− 1√

2 2√

2 6 20 2 2

The eigenvalues

of ATA are 8, 2, and 0, and since ATA issymmetric we know that the eigenvectors will be orthogonal. Thisgives eigenvectors

SVD. Examples

1√6− 1√

31√2

01√6− 1√

3− 1√

2 2√

2 6 20 2 2

The eigenvalues of ATA

are 8, 2, and 0, and since ATA issymmetric we know that the eigenvectors will be orthogonal. Thisgives eigenvectors

SVD. Examples

1√6− 1√

31√2

01√6− 1√

3− 1√

2 2√

2 6 20 2 2

The eigenvalues of ATA are 8, 2, and 0, and

since ATA issymmetric we know that the eigenvectors will be orthogonal. Thisgives eigenvectors

SVD. Examples

1√6− 1√

31√2

01√6− 1√

3− 1√

2 2√

2 6 20 2 2

The eigenvalues of ATA are 8, 2, and 0, and since ATA issymmetric

we know that the eigenvectors will be orthogonal. Thisgives eigenvectors

SVD. Examples

1√6− 1√

31√2

01√6− 1√

3− 1√

2 2√

2 6 20 2 2

The eigenvalues of ATA are 8, 2, and 0, and since ATA issymmetric we know that

the eigenvectors will be orthogonal. Thisgives eigenvectors

SVD. Examples

1√6− 1√

31√2

01√6− 1√

3− 1√

2 2√

2 6 20 2 2

The eigenvalues of ATA are 8, 2, and 0, and since ATA issymmetric we know that the eigenvectors will be orthogonal.

Thisgives eigenvectors

SVD. Examples

1√6− 1√

31√2

01√6− 1√

3− 1√

2 2√

2 6 20 2 2

SVD. Examples

1√63√121√12

02√6

SVD. Examples

1√63√121√12

02√6

SVD. Examples

1√63√121√12

02√6

SVD. Examples

1√63√121√12

02√6

SVD. Examples

1√63√121√12

02√6

SVD. Examples

1√63√121√12

02√6

SVD. Examples

1√63√121√12

02√6

SVD. Examples

3√12

0 −12

1√12− 2√

Finally, we can now we verify that we have A = UΣV T

A = UΣV T =1√6− 1√

31√2

01√6− 1√

3− 1√

2 00 0 0

3√12

0 −12

1√12− 2√

SVD. Examples

3√12

0 −12

1√12− 2√

A = UΣV T =1√6− 1√

31√2

01√6− 1√

3− 1√

2 00 0 0

3√12

0 −12

1√12− 2√

SVD. Examples

3√12

0 −12

1√12− 2√

A = UΣV T =1√6− 1√

31√2

01√6− 1√

3− 1√

2 00 0 0

3√12

0 −12

1√12− 2√

SVD. Examples

3√12

0 −12

1√12− 2√

Finally,

we can now we verify that we have A = UΣV T

A = UΣV T =1√6− 1√

31√2

01√6− 1√

3− 1√

2 00 0 0

3√12

0 −12

1√12− 2√

SVD. Examples

3√12

0 −12

1√12− 2√

A = UΣV T =1√6− 1√

31√2

01√6− 1√

3− 1√

2 00 0 0

3√12

0 −12

1√12− 2√

SVD. Examples

3√12

0 −12

1√12− 2√

A = UΣV T =

1√6− 1√

31√2

01√6− 1√

3− 1√

2 00 0 0

3√12

0 −12

1√12− 2√

SVD. Examples

3√12

0 −12

1√12− 2√

A = UΣV T =1√6− 1√

31√2

01√6− 1√

3− 1√

2 00 0 0

3√12

0 −12

1√12− 2√

linear algebra. session 9 - texas a&m universityroquesol/math_304_fall_2019_session_9.pdf ·...

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