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
Lb2
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 0.80.8 �0.6
�
A
�55
�=
�, A
76
�=
�, A
21
�=
�, A
�12
�=
�
Example:
⇒
⇒
An eigenvector of A corresponds to a unique eigenvalue. An eigenvalue of A has infinitely many eigenvectors.
A =
2
45 2 1�2 1 �12 2 4
3
5 v =
2
41�11
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
45 2 1�2 1 �12 2 4
3
5
2
41�11
3
5 =
2
44�44
3
5 = 4
2
41�11
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
43 0 00 1 20 2 1
3
5
2
40 1 �10 0 00 0 0
3
5
2
4x1
x2
x3
3
5 =
2
4x1
x3
x3
3
5 = x1
2
4100
3
5+ x3
2
4011
3
5
0 �11 0
�
7
T (x) = Ax = �xAn 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
