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• CHAPTER 5 EIGENVALUES, EIGENVECTORS, AND DIAGONALIZATION

Note: In these definitions v ∈ Rn and λ ∈ R, but sometimes it is necessary to extend the domain of T to allow v ∈ Cn and λ ∈ C.

Example: reflection operator T about the line y = (1/2)x

b1 is an eigenvector of T corresponding to the eigenvalue 1.

b2 is an eigenvector of T corresponding to the eigenvalue -1.

Definition

x

y

L b2

b1 = T (b1)

T (b2) = �b2

• Note: In these definitions v ∈ Rn and λ ∈ R, but sometimes it is necessary to allow v ∈ Cn and λ ∈ C.

Example:

non-eigenvectors eigenvectors

Definition

A =

 0.6 0.8 0.8 �0.6

A

 �5 5

� =

 � , A

 7 6

� =

 � , A

 2 1

� =

 � , A

 �1 2

� =

 �

• Example:

An eigenvector of A corresponds to a unique eigenvalue. An eigenvalue of A has infinitely many eigenvectors.

A =

2

4 5 2 1 �2 1 �1 2 2 4

3

5 v =

2

4 1 �1 1

3

5

Proof T(v) = λv ⇔ Av = λv.

The eigenvectors and corresponding eigenvalues of a linear operator are the same

as those of its standard matrix.

Property

Av =

2

4 5 2 1 �2 1 �1 2 2 4

3

5

2

4 1 �1 1

3

5 =

2

4 4 �4 4

3

5 = 4

2

4 1 �1 1

3

5 = 4v

A(cv) = c(Av) = c(4v) = 4(cv), 8c 2 R

• Proof Av = λv ⇔ Av - λv = 0 ⇔ Av - λInv = 0 ⇔ (A - λIn)v = 0.

Example: to check 3 and -2 are eigenvalues of the linear operator

Definition

Definition

T

✓ x1

x2

�◆ =

 �2x2

�3x1 + x2

Property Let A be an n⇥nmatrix with eigenvalue �. The eigenvectors of A corresponding to � are the nonzero solutions of (A� �In)v = 0.

• consider its standard matrix

The rank of A - 3I2 is 1, so its null space is not the zero space, and every nonzero vector in the null space is an eigenvector of T corresponding to the eigenvalue 3.

Similarly, the rank of A + 2I2 is 1, so its null space is not the zero space, and every nonzero vector in the null space is an eigenvector of T corresponding to the eigenvalue -2.

A =

 0 �2 �3 1

• Example: to check that 3 is an eigenvalue of B and find a basis for the corresponding eigenspace, where

find the reduced row echelon form of B - 3I3 and the solution set of (B - 3I3)x = 0, respectively, to be

Thus {[ 1 0 0 ]T, [ 0 1 1 ]T} is a basis of the eigenspace of B corresponding to the eigenvalue 3.

Example: some square matrices and linear operators on Rn have no

real eigenvalues, like the 90º-rotation matrix

B =

2

4 3 0 0 0 1 2 0 2 1

3

5

2

4 0 1 �1 0 0 0 0 0 0

3

5

2

4 x1

x2

x3

3

5 =

2

4 x1

x3

x3

3

5 = x1

2

4 1 0 0

3

5+ x3

2

4 0 1 1

3

5

 0 �1 1 0

• 7

T (x) = Ax = �x An eig. vec. x ===> unique eig. val. λ.

An eig. val. λ ===> multiple eig. vectors x. ===> The set of all eig. vectors corr. to λ is {x 2 Rn : (A� �In)x = 0} \ {0}

eigenspace corr. to the eig. val. λ

• Section 5.1: Problems 1, 3, 7, 9, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59

Homework Set for Sections 5.1