Eigenvalues, Eigenvectors, and Eigenspaces...
Transcript of Eigenvalues, Eigenvectors, and Eigenspaces...
Eigenvalues, Eigenvectors, and Eigenspaces
DEFINITION: Let A be a square matrix of size n.If a NONZERO vector ~x ∈ R
n and a scalar λ satisfy
A~x = λ~x, or, equivalently, (A − λIn)~x = 0,
scalar λ is called an eigenvalue of A,vector ~x 6= 0 is called an eigenvector of A associated with eigenvalue λ,and the null space of A − λI
nis called the eigenspace of A associated with eigenvalue λ.
HOW TO COMPUTE?
The eigenvalues of A are given by the roots of the polynomial
det(A − λIn) = 0.
The corresponding eigenvectors are the nonzero solutions of the linear system
(A − λIn)~x = 0.
Collecting all solutions of this system, we get the corresponding eigenspace.
EXERCISES:
For each given matrix, find the eigenvalues, and for each eigenvalue give a basis of thecorresponding eigenspace.
(a)
[
1 01 2
]
, (b)
[
2 01 2
]
, (c)
[
1 3−1 5
]
,
(d)
2 0 00 3 00 4 3
, (e)
−1 4 −2−3 4 0−3 1 3
, (f)
0 0 −21 2 11 0 3
, (g)
2 4 3−4 −6 −33 3 1
,
(h)
−1 1 1 −2−1 1 3 21 1 −1 −20 −1 −1 1
.
Turn over for the answers
Answers:
(a) Eigenvalues: λ1 = 1, λ2 = 2
The eigenspace associated to λ1 = 1, which is Ker(A − I): v1 =
[
−11
]
gives a basis.
The eigenspace associated to λ2 = 2, which is Ker(A − 2I): v2 =
[
01
]
gives a basis.
(b) Eigenvalues: λ1 = λ2 = 2
Ker(A − 2I), the eigenspace associated to λ1 = λ2 = 2: v1 =
[
01
]
gives a basis.
(c) Eigenvalues: λ1 = 2, λ2 = 4
Ker(A − 2I), the eigenspace associated to λ1 = 2: v1 =
[
31
]
gives a basis.
Ker(A − 4I), the eigenspace associated to λ2 = 4: v2 =
[
11
]
gives a basis.
(d) Eigenvalues: λ1 = 2, λ2 = λ3 = 3
Ker(A − 2I), the eigenspace associated to λ1 = 2: v1 =
100
gives a basis.
Ker(A − 3I), the eigenspace associated to λ2 = λ3 = 3: v2 =
001
gives a basis.
(e) Eigenvalues: λ1 = 1, λ2 = 2, λ3 = 3
The eigenspace associated to λ1 = 1: v1 =
111
gives a basis.
The eigenspace associated to λ2 = 2: v2 =
2/311
gives a basis.
The eigenspace associated to λ3 = 3: v3 =
1/43/41
gives a basis.
(f) Eigenvalues: λ1 = 1, λ2 = λ3 = 2
The eigenspace associated to λ1 = 1: v1 =
−211
gives a basis.
The eigenspace associated to λ2 = λ3 = 2: v2 =
010
, v3 =
−101
form a basis.
(g) Eigenvalues: λ1 = 1, λ2 = λ3 = −2
The eigenspace associated to λ1 = 1: v1 =
1−11
gives a basis.
The eigenspace associated to λ2 = λ3 = −2: v2 =
−110
gives a basis.
(h) Eigenvalues: λ1 = −1, λ2 = 1, λ3 = −2, λ4 = 2
The eigenspace associated to λ1 = −1: v1 =
1−110
gives a basis.
The eigenspace associated to λ2 = 1: v2 =
−11−11
gives a basis.
The eigenspace associated to λ3 = −2: v3 =
0−110
gives a basis.
The eigenspace associated to λ4 = 2: v4 =
−10−11
gives a basis.