chapter7_7.1_7.3_

7/29/2019 Chapter7_7.1_7.3_

1/53

1

7.1 Eigenvalues AndEigenvectors

7/29/2019 Chapter7_7.1_7.3_

2/53

2

Definition If A is an nn matrix, then a nonzero

vectorx in Rn is called an eigenvector

of A if Ax is a scalar multiple ofx; thatis,

Ax=x

for some scalar . The scalar is calledan eigenvalue of A, andx is said to bean eigenvector of Acorrespondingto .

7/29/2019 Chapter7_7.1_7.3_

3/53

3

Example 1

Eigenvector of a 22 Matrix

The vector is an eigenvectorof

Corresponding to the eigenvalue =3,since

1

2

x

3 0

8 1A

3 0 1 33

8 1 2 6A

x x

7/29/2019 Chapter7_7.1_7.3_

4/53

4

To find the eigenvalues of an nn matrix A werewrite Ax=x as

Ax=Ix

or equivalently,(I-A)x=0 (1)

For to be an eigenvalue, there must be a nonzerosolution of this equation. However, by Theorem 6.4.5,Equation (1) has a nonzero solution if and only if

det (I-A)=0This is called the characteristic equation of A; the scalar

satisfying this equation are the eigenvalues of A.When expanded, the determinant det (I-A) is apolynomial p in called the characteristic polynomial

of A.

7/29/2019 Chapter7_7.1_7.3_

5/53

5

Example 2

Eigenvalues of a 33 Matrix (1/3)

Find the eigenvalues of

Solution.The characteristic polynomial of A is

The eigenvalues of A must therefore satisfy the cubic equation

0 1 0

0 0 1

4 17 8

A

3 2

1 0

det( ) det 0 1 8 17 44 17 8

I A

3 28 17 4 0 (2)

7/29/2019 Chapter7_7.1_7.3_

6/53

6

Example 2

Eigenvalues of a 33 Matrix (2/3)

To solve this equation, we shall begin by searching forinteger solutions. This task can be greatly simplifiedby exploiting the fact that all integer solutions (if

there are any) to a polynomial equation with integercoefficientsn+c1

n-1++cn=0must be divisors of the constant term cn. Thus, the only

possible integer solutions of (2) are the divisors of -4,

that is,1,

2,

4. Successively substituting thesevalues in (2) shows that 4 is an integer solution.

As a consequence, -4 must be a factor of the leftside of (2). Dividing -4 into 3-82+17-4 show that(2) can be rewritten as

(-4)(2-4+1)=0

7/29/2019 Chapter7_7.1_7.3_

7/53

7

Example 2

Eigenvalues of a 3

3 Matrix (3/3)Thus, the remaining solutions of (2) satisfy the

quadratic equation

2-4+1=0which can be solved by the quadratic formula.

Thus, the eigenvalues of A are

4, 2 3, 2 3

7/29/2019 Chapter7_7.1_7.3_

8/53

8

Example 3Eigenvalues of an Upper Triangular

Matrix (1/2) Find the eigenvalues of the upper triangular matrix

Solution.

Recalling that the determinant of a triangular matrix isthe product of the entries on the main diagonal(Theorem 2.2.2), we obtain

11 12 13 14

22 23 24

33 34

44

00 0

0 0 0

a a a a

a a aA

a a

a

l

7/29/2019 Chapter7_7.1_7.3_

9/53

9

Example 3Eigenvalues of an Upper Triangular

Matrix (2/2)

Thus, the characteristic equation is

(-a11)(-a22) (-a33) (-a44)=0and the eigenvalues are

=a11, =a22, =a33, =a44which are precisely the diagonal entries of A.

11 12 13 14

22 23 24

33 34

44

11 22 33 44

0det( ) det

0 0

0 0 0

( )( )( )( )

a a a a

a a aI A

a a

a

a a a a

7/29/2019 Chapter7_7.1_7.3_

10/53

10

Theorem 7.1.1 If A is an nn triangular matrix (upper

triangular, low triangular, or diagonal),

then the eigenvalues of A are entries onthe main diagonal of A.

E l 4

7/29/2019 Chapter7_7.1_7.3_

11/53

11

Example 4Eigenvalues of a Lower Triangular

Matrix By inspection, the eigenvalues of the lower

triangular matrix

are =1/2, =2/3, and =-1/4.

1/ 2 0 01 2 / 3 0

5 8 1/ 4

A

7/29/2019 Chapter7_7.1_7.3_

12/53

12

Theorem 7.1.2

Equivalent Statements If A is an nn matrix and is a real number,

then the following are equivalent.

a) is an eigenvalue of A.b) The system of equations (I-A)x=0 has

nontrivial solutions.

c) There is a nonzero vectorx in Rn such that

Ax=x.d) is a solution of the characteristic equation

det(I-A)=0.

7/29/2019 Chapter7_7.1_7.3_

13/53

13

Finding Bases for Eigenspaces The eigenvectors of A corresponding to

an eigenvalue are the nonzerox that

satisfy Ax=x. Equivalently, theeigenvectors corresponding to are thenonzero vectors in the solution space of

(I-A)x=0. We call this solution spacethe eigenspaceof A corresponding to .

7/29/2019 Chapter7_7.1_7.3_

14/53

14

Example 5

Bases for Eigenspaces (1/5) Find bases for the eigenspaces of

Solution.

The characteristic equation of matrix A is 3-52+8-4=0,or in factored form, (-1)(-2)2=0; thus, theeigenvalues of A are =1 and =2, so there are twoeigenspaces of A.

0 0 2

1 2 1

1 0 3

A

7/29/2019 Chapter7_7.1_7.3_

15/53

15

Example 5

Bases for Eigenspaces (2/5)By definition,

Is an eigenvector of A corresponding to if and only ifxis a nontrivial solution of (I-A)x=0, that is, of

If =2, then (3) becomes

1

2

3

x

x

x

x

1

2

3

0 2 01 2 1 0 (3)

1 0 3 0

x

x

x

7/29/2019 Chapter7_7.1_7.3_

16/53

16

Example 5Bases for Eigenspaces (3/5)

Solving this system yieldx1=-s, x2=t, x3=s

Thus, the eigenvectors of A corresponding to =2 are thenonzero vectors of the form

Since

1

2

3

2 0 2 0

1 0 1 0

1 0 1 0

x

x

x

0 1 0

0 0 1

0 1 0

s s

t t s t

s s

x

1 0

0 and 1

1 0

7/29/2019 Chapter7_7.1_7.3_

17/53

17


are linearly independent, these vectors form a basis forthe eigenspace corresponding to 2.

If 1, then (3) becomes

Solving this system yieldsx1=-2s, x2=s, x3=s

1

2

3

1 0 2 0

1 1 1 0

1 0 2 0

x

x

x

7/29/2019 Chapter7_7.1_7.3_

18/53

18


Thus, the eigenvectors corresponding to 1

are the nonzero vectors of the form

is a basis for the eigenspace corresponding to1.

2 2 -2

1 so that 1

1 1

s

s s

s

7/29/2019 Chapter7_7.1_7.3_

19/53

19

Theorem 7.1.3

Ifkis a positive integer, is aneigenvalue of a matrix A, andx is

corresponding eigenvector, then k is aneigenvalue of Akandx is acorresponding eigenvector.

7/29/2019 Chapter7_7.1_7.3_

20/53

20

Example 6Using Theorem 7.1.3 (1/2)

In Example 5 we showed that the eigenvalues of

are =2 and =1, so from Theorem 7.1.3 both =27=128and =17=1 are eigenvalues of A7. We also showed that

are eigenvectors of A corresponding to the eigenvalue =2,so from Theorem 7.1.3 they are also eigenvectors of A7corresponding to =27=128. Similarly, the eigenvector

0 0 2

1 2 1

1 0 3

A

1 0

0 and 1

1 0

7/29/2019 Chapter7_7.1_7.3_

21/53

21

Example 6Using Theorem 7.1.3 (2/2)

of A corresponding to the eigenvalue =1 isalso eigenvector of A7 corresponding to

=17=1.

2

1

1

7/29/2019 Chapter7_7.1_7.3_

22/53

22

Theorem 7.1.4

A square matrix A is invertible if andonly if =0 is not an eigenvalue of A.

7/29/2019 Chapter7_7.1_7.3_

23/53

23

Example 7Using Theorem 7.1.4

The matrix A in Example 5 is invertiblesince it has eigenvalues =1 and =2,

neither of which is zero. We leave it forreader to check this conclusion byshowing that det(A)0

7/29/2019 Chapter7_7.1_7.3_

24/53

24

Theorem 7.1.5Equivalent Statements (1/3)

If A is an nn matrix, and if TA: RnRn is

multiplication by A, then the following areequivalent.

a) A is invertible.b) Ax=0 has only the trivial solution.c) The reduced row-echelon form of A is In.d) A is expressible as a product of elementary matrix.e) Ax=b is consistent for every n1 matrix b.f) Ax=b has exactly one solution for every n1 matrix

b.g) det(A)0.

7/29/2019 Chapter7_7.1_7.3_

25/53

25


h) The range of TA is Rn.

i) TA is one-to-one.j)

The column vectors of A are linearlyindependent.k) The row vectors of A are linearly

independent.l) The column vectors of A span Rn.m) The row vectors of A span Rn.n) The column vectors of A form a basis for Rn.o) The row vectors of A form a basis for Rn.

7/29/2019 Chapter7_7.1_7.3_

26/53

26


p) A has rank n.

q) A has nullity 0.

r) The orthogonal complement of thenullspace of A is Rn.

s) The orthogonal complement of the

row space of A is {0}.t) ATA is invertible.

u) =0 is not eigenvalue of A.

7/29/2019 Chapter7_7.1_7.3_

27/53

27

7.2 Diagonalization

7/29/2019 Chapter7_7.1_7.3_

28/53

28

Definition

A square matrix A is calleddiagonalizable if there is an invertible

matrix P such that P-1AP is a diagonalmatrix; the matrix P is said todiagonalize A.

7/29/2019 Chapter7_7.1_7.3_

29/53

29

Theorem 7.2.1

If A is an nn matrix, then the

following are equivalent.

a) A is diagonalizable.

b) A has n linearly independenteigenvectors.

7/29/2019 Chapter7_7.1_7.3_

30/53

30

Procedure for Diagonalizing aMatrix

The preceding theorem guarantees that an nnmatrix A with n linearly independent eigenvectors isdiagonalizable, and the proof provides the followingmethod for diagonalizing A.

Step 1. Find n linear independent eigenvectors of A,say, p1, p2, , pn.

Step 2. From the matrix P having p1, p2, , pn as itscolumn vectors.

Step 3. The matrix P-1

AP will then be diagonal with 1,2, , nas its successive diagonal entries, where iis the eigenvalue corresponding to pi, for i=1, 2, ,n.

Example 1

7/29/2019 Chapter7_7.1_7.3_

31/53

31

Example 1Finding a Matrix P That Diagonalizesa Matrix A (1/2)

Find a matrix P that diagonalizes

Solution.From Example 5 of the preceding section we found the

characteristic equation of A to be(-1)(-2)2=0

and we found the following bases for the eigenspaces:

0 0 2

1 2 1

1 0 3

A

1 2 3

1 0 2

2 : 0 , 1 =1: 1

1 0 1

p p p

Example 1

7/29/2019 Chapter7_7.1_7.3_

32/53

32

Example 1Finding a Matrix P That Diagonalizesa Matrix A (2/2)

There are three basis vectors in total, so the matrix A isdiagonalizable and

diagonalizes A. As a check, the reader should verify

that

1 0 2

0 1 1

1 0 1

P

1

1 0 2 0 0 2 1 0 2 2 0 0

1 1 1 1 2 1 0 1 1 0 2 0

1 0 1 1 0 3 1 0 1 0 0 1

P AP

Example 2

7/29/2019 Chapter7_7.1_7.3_

33/53

33

Example 2A Matrix That Is Not Diagonalizable(1/4)

Find a matrix P that diagonalize

Solution.

The characteristic polynomial of A is

1 0 0

1 2 0

3 5 2

A

2

1 0 0

det( ) 1 2 0 ( 1)( 2)

3 5 2

I A

Example 2

7/29/2019 Chapter7_7.1_7.3_

34/53

34


so the characteristic equation is

(-1)(-2)2=0

Thus, the eigenvalues of A are =1 and =2. We leaveit for the reader to show that bases for theeigenspaces are

Since A is a 33 matrix and there are only two basis

vectors in total, A is not diagonalizable.

1 2

1/ 8 0

1: 1/ 8 2 : 01 1

p p

Example 2

7/29/2019 Chapter7_7.1_7.3_

35/53

35


Alternative Solution.If one is interested only in determining whether a matrix

is diagonalizable and is not concerned with actuallyfinding a diagonalizing matrix P, then it is not

necessary to compute bases for the eigenspaces; itsuffices to find the dimensions of the eigenspaces.For this example, the eigenspace corresponding to=1 is the solution space of the system

The coefficient matrix has rank 2. Thus, the nullity ofthis matrix is 1 by Theorem 5.6.3, and hence the

solution space is one-dimensional.

1

2

3

0 0 0 0

1 1 0 0

3 5 1 0

x

x

x

Example 2

7/29/2019 Chapter7_7.1_7.3_

36/53

36


The eigenspace corresponding to =2 is thesolution space system

This coefficient matrix also has rank 2 and nullity 1,

so the eigenspace corresponding to =2 is alsoone-dimensional. Since the eigenspaces producea total of two basis vectors, the matrix A is notdiagonalizable.

1

2

3

1 0 0 0

1 0 0 0

3 5 0 0

x

x

x

7/29/2019 Chapter7_7.1_7.3_

37/53

37

Theorem 7.2.2

Ifv1, v2, vk, are eigenvectors of A

corresponding to distinct eigenvalues 1,

2, , k, then{v1, v2, vk} is alinearly independent set.

7/29/2019 Chapter7_7.1_7.3_

38/53

38

Theorem 7.2.3

If an nn matrix A has n distinct

eigenvalues, then A is diagonalizable.

7/29/2019 Chapter7_7.1_7.3_

39/53

39

Example 3Using Theorem 7.2.3

We saw in Example 2 of the preceding section that

has three distincteigenvalues, . Therefore, Ais diagonalizable. Further,

for some invertible matrix P. If desired, the matrix P canbe found using method shown in Example 1 of thissection.

0 1 0

0 0 1

4 17 8

A

4, 2 3, 2 3

1

4 0 0

0 2 3 0

0 0 2 3

P AP

7/29/2019 Chapter7_7.1_7.3_

40/53

40

Example 4A Diagonalizable Matrix

From Theorem 7.1.1 the eigenvalues of atriangular matrix are the entries on its maindiagonal. This, a triangular matrix withdistinct entries on the main diagonal isdiagonalizable. For example,

is a diagonalizable matrix.

1 2 4 0

0 3 1 7

0 0 5 8

0 0 0 2

A

7/29/2019 Chapter7_7.1_7.3_

41/53

41

Theorem 7.2.4Geometric and Algebraic Multiplicity

If A is a square matrix, then :

a) For every eigenvalue of A the

geometric multiplicity is less than orequal to the algebraic multiplicity.

b) A is diagonalizable if and only if thegeometric multiplicity is equal to thealgebraic multiplicity for everyeigenvalue.

7/29/2019 Chapter7_7.1_7.3_

42/53

42

Computing Powers of a Matrix(1/2)

There are numerous problems in appliedmathematics that require the computation ofhigh powers of a square matrix. We shall

conclude this section by showing howdiagonalization can be used to simplify suchcomputations for diagonalizable matrices.

If A is an nn matrix and P is an invertiblematrix, then

(P-1AP)2=P-1APP-1AP=P-1AIAP=P-1A2PMore generally, for any positive integer k

(P-1AP)k=P-1AkP (8)

7/29/2019 Chapter7_7.1_7.3_

43/53

43

Computing Powers of a Matrix(2/2)

It follows form this equation that if A is diagonalizable,and P-1AP=D is a diagonal matrix, then

P-1AkP=(P-1AP)k=Dk (9)

Solving this equation for Ak

yieldsAk=PDkP-1 (10)

This last equation expresses the kth power of A in termsof the kth power of the diagonal matrix D. But Dk iseasy to compute; for example, if

1 1

2 k 2

0 ... 0 0 ... 0

0 ... 0 0 ... 0, and D

: : : : : :

0 0 ... 0 0 ...

k

k

k

n n

d d

d dD

d d

7/29/2019 Chapter7_7.1_7.3_

44/53

44

Example 5Power of a Matrix (1/2)

Using (10) to find A13, where

Solution.

We showed in Example 1 that the matrix A is diagonalized by

and that

0 0 2

1 2 1

1 0 3

A

1 0 2

0 1 1

1 0 1

P

1

2 0 0

0 2 0

0 0 1

D P AP

7/29/2019 Chapter7_7.1_7.3_

45/53

45

Example 5Power of a Matrix (2/2)

Thus, form (10)

13

13 13 1 13

13

1 0 2 2 0 0 1 0 2

0 1 1 0 2 0 1 1 1

1 0 1 0 0 2 1 0 1

8190 0 16382

8191 8192 8191 (13)8191 0 16383

A PD P

7/29/2019 Chapter7_7.1_7.3_

46/53

46

7.3 OrthogonalDiagonalization

7/29/2019 Chapter7_7.1_7.3_

47/53

47

The OrthogonalDiagonalization Matrix Form

Given an nn matrix A, if there exist an

orthogonal matrix P such that the

matrix P-1AP=PTAP, then A is said to beorthogonally diagonalizable and P issaid to orthogonally diagonalize A.

7/29/2019 Chapter7_7.1_7.3_

48/53

48

Theorem 7.3.1

If A is an nn matrix, then the

following are equivalent.

a) A is orthogonally diagonalizable.

b) A has an orthonormal set of neigenvectors.

c) A is symmetric.

7/29/2019 Chapter7_7.1_7.3_

49/53

49

Theorem 7.3.2

If A is a symmetric matrix, then:

a) The eigenvalues of A are real

numbers.

b) Eigenvectors from differenteigenspaces are orthogonal.

7/29/2019 Chapter7_7.1_7.3_

50/53

50

Diagonalization of SymmetricMatrices

As a consequence of the preceding theoremwe obtain the following procedure fororthogonally diagonalizing a symmetric matrix.

Step 1. Find a basis for each eigenspace of A.Step 2. Apply the Gram-Schmidt process to

each of these bases to obtain an orthonormalbasis for each eigenspace.

Step 3. Form the matrix P whose columns arethe basis vectors constructed in Step2; thismatrix orthogonally diagonalizes A.

Example 1

7/29/2019 Chapter7_7.1_7.3_

51/53

51

Example 1An Orthogonal Matrix P ThatDiagonalizes a Matrix A (1/3)

Find an orthogonal matrix P that diagonalizes

Solution.

The characteristic equation of A is

4 2 2

2 4 2

2 2 4

A

2

4 2 2

det( ) det 2 4 2 ( 2) ( 8) 0

2 2 4

I A

Example 1

7/29/2019 Chapter7_7.1_7.3_

52/53

52


Thus, the eigenvalues of A are =2 and =8. By themethod used in Example 5 of Section 7.1, it can beshown that

form a basis for the eigenspace corresponding to =2.Applying the Gram-Schmidt process to {u1, u2}

yields the following orthonormal eigenvectors:

1 2

1 1

1 and 0

0 1

u u

1 2

1/ 2 1/ 6

1/ 2 and 1/ 6

0 2 / 6

v v

Example 1

7/29/2019 Chapter7_7.1_7.3_

53/53


The eigenspace corresponding to =8 has

as a basis. Applying the Gram-Schmidt process to {u3} yields

Finally, using v1, v2, and v3 as column vectors we obtain

3

1

1

1

u

3

1/ 3

1/ 3

1/ 3

v

1/ 2 1/ 6 1/ 3

1/ 2 1/ 6 1/ 3

0 2 / 6 1/ 3

P

chapter7_7.1_7.3_

Documents