1e_matrixeigenproblemsnotes

7/29/2019 1e_MatrixEigenProblemsNotes

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EMT451T/MMA401T: EigenProblemRevision Tshwane University of Technology: EL Voges 1 of 13

REVISION: Eigen problems

(Abstract from Mathematics III notes)

All you need to know for EMT451T/MMA401T is:

How to calculate eigenvalues How to calculate eigenvectors associated with distinct and repeating eigenvalues

complex eigenvalues excluded!

No solution of ODEs!

1. DEFINITIONS AND THEOREMS

Consider the general system of n n first order, linear ordinary differential equationswith constant coefficients

111 1 12 2 1 1

221 1 22 2 2 2

1 1 2 2

( )

( )

( ).

n n

n n

nn n nn n n

dxa x a x a x f t

dt

dxa x a x a x f t

dt

dxa x a x a x f t

dt

(1)

Define matricesX,A andF(t) as

1

2

( )

( )

( )n

x t

x tX

x t

,

11 12 1

21 22 2

1 2

( )

n

n

n n nn

a a a

a a aA t

a a a

and

1

2

( )

( )( )

( )n

f t

f tF t

f t

.

System (1) may the be written in matrix form as

1

11 12 1 1 1

221 22 2 2 2

1 2

( ) ( )

( ) ( )

( ) ( )

n

n

n n nn n nn

dxdt a a a x t f t

dxa a a x t f t

dt

a a a x t f t dx

dt


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or

'X AX F .

When ( ) 0F t , that is, ( ) 0if t for all values ofi, this system is said to be

homogeneous; otherwise it is non-homogeneous. In this study unit we are interested inthe homogeneous case only. We thus consider the system of equations represented inmatrix form as

'X AX . (2)

NB!

Remember, here

0

00 0 0 0

0

T

, the 1n zero column vector.

A solution of the system (2) on an intervalIis any column vectorX, called a solution

vector, with differentiable functions as entries where,

1

2

( )

( )( )

( )n

x t

x tX t

x t

. (3)

EXAMPLE 1

Show that6

3

5

tX e

is a solution of the matrix equation1 3

'5 3

X X

on the interval

; .SOLUTION

From6

3

5

tX e

we know that6

3' 6

5

tX e

.

Thus,

6 6 61 3 1 3 3 18 3

6 '5 3 5 3 5 30 5

t t tX e e e X

.

63

5

tX e

is a solution of1 3

'5 3

X X

.


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Recall that with the method of undetermined coefficients we assumed (Singh, 2003:657)

the solution of a single linear ordinary differential equation2

2( ), 0

d y dya b cy f x a

dx dx

with constant coefficients is of the form

mty Ae .

In a similar way we assume the solution of the system of linear ordinary differential

equation with constant coefficients (2) has the form

1

2 t t

n

k

kX e Ke

k

(4)

with some constant. Differentiation of the solution (4) yields

't

X Ke

. (5)

Thus, if the vector (4) is the solution of the system (2), then substitution of (4) and (5) in

(2) results int tKe AKe .

Division byte leads to

AK K (6)and rearrangement of terms yields

0AK K or

( ) 0A I K (7)

whereI is the n n identity matrix. From (7), either 0K , which is the trivial solutionand of no further interest, or

0A I . (8)

Definition 1 Characteri stic equation and polynomial , eigenvalue and eigenvector.Equation (8) is called the characteristic equation of matrixA. The solutions of the

characteristic equation are called the eigenvalues ofA. The nth degree polynomialA I is the characteristic polynomial. A solution 0K of Equation (7)

corresponding to a specific eigenvalue is called an eigenvector. Furthermore,tX Ke

is a solution of the system (2).


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NB!The word "eigenvalue" was derived from the German word "eigenwert" which literally

translates as "proper value" (Zill, 2000:395). Eigenvalues and eigenvectors are sometimescalled characteristic values and characteristic vectors respectively.

NB!Note that 0K , that is, an eigenvectorcannot be a zero vector. However, 0 can be

an eigenvalue.

NB!In the characteristic equation, Equation (8), we use the determinant, 0A I , to

calculate the eigenvalues.

NB!We use Equation (7), ( ) 0A I K , containing the matrix ( )A I to calculate the

eigenvectors.

NB!In this study unit it will be possible to factorize the characteristic polynomial or use theso-called quadratic formula to calculate the eigenvalues. However, in real-life problems

we would probably use numerical methods to determine the eigenvalues.

NB!You may come across leftand righteigenvectors in the literature. Good news

our studyis limited to left eigenvectors only. We therefore always refer to eigenvectors implying

left eigenvectors.

Figure 1 depicts the influence of on an eigenvectorK. 1 represents "stretching",0 1 represents "shrinking", and 0 represents change in direction.

Fi ure 1AK= Kfor different values of

0< 1


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EXAMPLE 2

Show that1

1K

is an eigenvector of

2 3

1 2A

.

SOLUTION

From Equation (6) we know thatKis an eigenvector ofA ifAK K :

eigenvalue

2 3 1 1 11

1 2 1 1 1AK K

.

Thus,Kis an eigenvector ofA.

NB!Note that the eigenvalue in this case is -1.

NB!Remember, for matricesA andK, AK KA , except in special cases.

When solving algebraic equations we usually get more than one possible answer. In a

similar way we get more than one solution vector, called a set of solution vectors, when

we solve a system of first order, linear ordinary differential equations with constant

coefficients. We now look at a few definitions and theorems regarding these solutionvectors.

NB!Proofs of theorems are not included; you may find them in most of the books listed at the

end of this study unit.

Theorem 1 The superposit ion pr inciple

Let 1 2; ; nX X X be a set of solution vectors of the homogeneous system of equations(2) on an intervalI. Then the linear combination

1 1 2 2 n nX c X c X c X ,

with , 1; 2; ;ic i n arbitrary constants, is also a solution on the interval.

Definition 2L inear dependence/independence

Let 1 2; ; nX X X be a set of solution vectors of the homogeneous system of equations(2) on an intervalI. The set is linearly dependent on the interval if there exist constants

1 2, , , kc c c , not all zero, such that

1 1 2 2 0k kc X c X c X (9)

for every ton the interval. If the set of solution vectors is not linearly dependent, then it is

linearly independent.


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Let's visualize the concept of linear dependence/independence. Consider the following

vectors depicted in Figure 2:

1 2 3

1 2 4; ;

2 4 1K K K

.

Vectors 1K and 2K have the same direction, while 3K has a different direction. Thus, 1K

and 2K are linearly dependent. 3K and 1K are linearly independent.

EXAMPLE 3

Are the vectors 14

10X

and 2

2

5X

linearly independent? Motivate your answer.

SOLUTION

From Equation (9) we know the two vectors will be linearly dependent if we can

find some constant 0c such that

1 20X cX ,

that is, if

1 2X cX .

But

1 2

4 22 2

10 5X X

.

Thus, there exists 2c for which 1 2 0X cX .

X1 andX2are not linearly independent, that is, they are linearly dependent.

Figure 2 Linear dependent/independent vectors

K

K

K

3

2

1

(4;1)

(1;2)

(2;4)


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NB!From the example above we see that vectors are linearly dependent if they are constant

multiples of each other.

The method illustrated in Example 3 may become tedious when we want to prove linearindependence of solution vectors. The next definition and theorem gives us a "simple

tool" to use.

Definition 3 WronskianThe Wronskian of a set of vectors

11 12 1

21 22 2

1 2

1 2

, , ,

n

n

n

n n nn

x x x

x x xX X X

x x x

is the real-valued function

11 12 1

21 22 2

1 2

1 2

[ , , , ]

n

n

n

n n nn

x x x

x x xW X X X

x x x

.

NB!

The solution vectors form the columns in the determinant.

Theorem 2 Cri teri on for l inear i ndependent solutionsLet

11 12 1

21 22 2

1 2

1 2

, , ,

n

n

n

n n nn

x x x

x x xX X X

x x x

be a n solution vectors of the homogeneous system (2) on an intervalI. Then the set of

solution vectors is linearly independent if and only if the Wronskian

11 12 1

21 22 2

1 2

1 2

[ , , , ] 0

n

n

n

n n nn

x x x

x x xW X X X

x x x

for at least one value ofton the interval.


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EXAMPLE 4

Show that the 13

1

tX e

and 21

1

tX e

are linearly independent on the interval

; .

SOLUTION

We know that 13 3

1

t

t

t

eX e

e

and 2

1

1

t

t

t

eX e

e

.

1 23

[ , ] 3 3 1 2t t

t t t t

t t

e eW X X e e e e

e e

.

Thus, 1 2[ , ] 0W X X for all values oftandX1 andX2 are linearly independent.

NB!You may come across a factor

at

e in the Wronskian. Remember, 0at

e for all realvalues ofa and t.

Definition 4 Fundamental set of solut ions

Any set 1 2; ; nX X X of linearly independent solution vectors of the system (2) onsome intervalIis called a fundamental set of solutions on the interval.

Theorem 3 Existence of a fundamental set of solutionsThere exist a set of fundamental solutions for system (2) on some intervalI.

Theorem 4 General solu tion of a homogeneous system

Let 1 2; ; nX X X be a fundamental set of solution vectors of the system (2) on someintervalI. Then the general solution of the system on the interval is

1 1 2 2 n nX c X c X c X

where , 1, 2, ,ic i n are arbitrary constants.

EXAMPLE 5

Show that the solution vectors 13

1tX e

and 21

1tX e

form a fundamental set of

solutions for2 3

'1 2

X X

.

SOLUTION

In the previous example we showed thatX1 andX2 are linearly independent. Thus,

from Definition 4, the set 1 2,X X is a fundamental set of solutions.


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2. EIGENVALUES AND EIGENVECTORS

In the previous activity we came across three different types of eigenvalues: real and

distinct eigenvalues, complex eigenvalues, and real and repeating eigenvalues. There areslight differences in the techniques used to calculate the eigenvectors associated with

each type of eigenvalue. We shall, for each type of eigenvalue, first calculate the

corresponding eigenvectors and then look at the solutions of systems of differentialequations.

We shall use Equations (7) and (8) to calculate eigenvalues and eigenvectors. To enhance

reading we copied the equations here. Eigenvalues, , are calculated from

0A I (10)

and the eigenvectors,K, from

0A I K . (11)

2.1 REAL AND DISTICT EIGENVALUES

2.1.1 THE CORRESPONDING EIGENVECTORS

Let's illustrate the process in detail using the matrix2 3

1 2A

as an example.

Using Equation (10),

2 3 1 0 2 3 0 2 3

1 2 0 1 1 2 0 1 2

A I

.

The characteristic equation is therefore

2 30

1 2

NB!Note the position of the on the principle diagonal. This is typical of the characteristic

equation; in future we are going to write this equation down directly.

Evaluation of the determinant yields

(2 )( 2 ) 3 0 which simplifies to2 1 0

with roots

1 21; 1


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EMT451T/MMA401T: EigenProblemRevision Tshwane University of Technology: EL Voges10 of 13

To calculate the eigenvector associated with1 1 , substitute in Equation (11) and

solve forK:

1

2

2 1 3 0

1 2 1 0

k

k

,

that is,1

2

1 3 0

1 3 0

k

k

.

NB!Note that the two rows in the matrix on the left are exactly the same. We thus have one

equation in two unknowns to solve.

Writing this matrix equation as separate equations yield one equation

1 23 0k k .Thus,

1 23k k

where2k is any real number, say a. Then 1 3k a and the general eigenvector associated

with 1 1 is of the form

1

3aK

a

.

The a may be any real number. If we set 2 1k , then 1 3k and the associated or

corresponding eigenvector is

131

K

.

NB!

What if we choose 2 2k ? Then 1 6k and*

6

2K

. Do we now have two different

eigenvectors associated with the same eigenvalue? No! Since * 12K K these two vectors

are not linearly independent; we can use either one of them, but not both.

NB!For every eigenvalue there are infinitely many general eigenvectors. We may choose any

one of those eigenvectors; we usually choose 2 1k .

For 2 1 the matrix equation (11) is


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1

2

2 ( 1) 3 0

1 2 ( 1) 0

k

k

or

1

2

3 3 0

1 1 0

k

k

.

NB!Note that the two rows are multiples of each other. We thus, once again, have oneequation in two unknowns.

NB!Which row should we use to calculate the k's? It doesn't really matter, but it is logical to

choose the row leading to the simplest calculations.

Using the second row,

1 2 0k k ,

that is,

1 2k k

where k2 is any real number. Let 2 1k , then 1 1k and

2

1

1K

.

Thus, eigenvector 1

3

1K

corresponds to eigenvalue1 1 , and

2

1

1K

to2 1 .

NB!We cannot choose

2 0k . Why not? If 2 0k , then 1 0k and the eigenvector is

1 0 0T

K , which is, per definition, not possible.

NB!We usually choose the variable on the right, in this case 2k , equal to one.

NB!Here is a "nice" property of distinct, real eigenvalues you can use to check your

eigenvalues: Trace ofA = sum of eigenvalues. In this example, trace 2 2 0A and

1 2 1 1 0 . (Trace = sum of entries on principal diagonal.)


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EXAMPLE 6

Determine the eigenvalues and associated eigenvectors of4 2

5 1

.

SOLUTION

Characteristic equation:

4 20

5 1

.

Evaluating the determinant,

(4 )(1 ) 10 0 which simplifies to

2 5 6 0 .Factorization yields

( 1)( 6) 0

so that

1 1 and 2 6 .

1

2

5 2 0

5 2 0

k

k

1

2

2 2 0

5 5 0

k

k

1 25 2 0k k 1 22 2 0k k 2

1 25k k 1 2k k

Let 22 1 51k k Let 2 11 1k k 2

5

11

K

21

1K

.

EXAMPLE 7Calculate the eigenvalues and associated eigenvectors of

1 2 3

0 5 6

0 0 7

A

.

SOLUTIONThe characteristic equation is

1 2 3

0 5 6 0

0 0 7

which yields

(1 )(5 )( 7 ) 0 so that

1 2 31; 5; 7 .


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1

1 2

3

0 2 3 0

1 0 4 6 0

0 0 8 0

k

k

k

.

From the last row,3 38 0 0k k .

From the first row 2 3 22 3 0 0k k k .

The second row is a constant multiple of the first; how do we calculate 1k

? Here 1k may be any real number and we choose 1 1k . Thus, 1

1

0

0

K

.

1

2 2

3

4 2 3 05 0 0 6 0

0 0 12 0

kk

k

.

From both the second and the last row 3 0k .

From the first row 11 2 1 224 2 0 0k k k k .

Let 12 1 21k k . Thus,

12

2 1

0

K

.

1

3 2

3

8 2 3 0

7 0 12 6 0

0 0 0 0

k

k

k

From the second row 12 3 2 3212 6 0k k k k .

From the first row 2 31 2 3 12 3

8 2 3 08

k kk k k k

.

Let 1 13 2 12 41 ;k k k . Thus,

14

123

1

K

.

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