END 210E Linear Algebra for Industrial Engineers
Worked Examples: Eigenvalues and Eigenvectors
2024-25 Spring
1. Find the characteristic equation of the following symmetric matrices, and then by inspection, determine
the dimensions of the eigenspaces:
1
a. [
2
2
]
4
1 1 1
b. [1 1 1]
1 1 1
4
4
c. [
0
0
a.
4
4
0
0
0
0
0
0
0
0
]
0
0
The characteristic equation:
det ( [
1
2
2
1
]−𝜆[
4
0
0
1−𝜆
] ) = det ( [
1
2
2
] ) = 𝜆2 − 5𝜆 = 𝜆(𝜆 − 5) = 0.
4−𝜆
Then, the eigenspace for the eigenvalue 𝜆 = 0 is one-dimensional, and the eigenspace for the eigenvalue
𝜆 = 5 is also one-dimensional.
b. The characteristic equation:
1−𝜆
det ( [ 1
1
1
1−𝜆
1
1
1 ] ) = 𝜆3 − 3𝜆2 = 𝜆2 (𝜆 − 3) = 0.
1−𝜆
Then, the eigenspace for the eigenvalue 𝜆 = 3 is one-dimensional, and the eigenspace for the eigenvalue
𝜆 = 0 is two-dimensional.
c.
The characteristic equation:
4−𝜆
4
det ( [
0
0
4
4−𝜆
0
0
0
0
−𝜆
0
0
0
] ) = 𝜆4 − 8𝜆3 = 𝜆3 (𝜆 − 8) = 0.
0
−𝜆
Then, the eigenspace for the eigenvalue 𝜆 = 0 is three-dimensional, and the eigenspace for the eigenvalue
𝜆 = 8 is one-dimensional.
2 −1 −1
2. Find a matrix 𝑷 that diagonalizes 𝑨 = [−1
2 −1], and determine 𝑷−𝟏 𝑨𝑷.
−1 −1
2
We should first find the eigenvalues and the corresponding eigenvectors:
2−𝜆
det ([ −1
−1
−1
2−𝜆
−1
−1
−1 ]) = (𝜆 − 3)2 𝜆 = 0 ⇒ 𝜆1 = 𝜆2 = 3, 𝜆3 = 0.
2−𝜆
−1
When 𝜆 = 3, 𝑨𝒗 = 3𝒗 ⇒ (𝑨 − 3𝑰)𝒗 = 𝟎 ⇒ [−1
−1
−1
−1
−1
0
−1 𝑣1
−1] [𝑣2 ] = [ 0 ] ⇒ 𝑣1 + 𝑣2 + 𝑣3 = 0.
−1 𝑣3
0
𝑣1
−𝑠 − 𝑡
−1
−1
𝑣
For 𝑣2 = 𝑠 ∈ ℝ and 𝑣3 = 𝑡 ∈ ℝ, 𝑣1 = −𝑠 − 𝑡. Then, 𝒗 = [ 2 ] = [ 𝑠 ] = 𝑠 [ 1 ] + 𝑡 [ 0 ].
𝑣3
𝑡
0
1
−1
−1
The corresponding eigenspace is the plane spanned by [ 1 ] and [ 0 ].
0
1
2
When 𝜆 = 0, 𝑨𝒗 = 0𝒗 ⇒ 𝑨𝒗 = 𝟎 ⇒ [−1
−1
1
1
⇒ [−1
2
0 −3
1 0
⇒[ 0 1
0 0
−2
−1
3
−1
−1
0
⋮
⋮
⋮
⋮
⋮
⋮
0
1
0] ⇒ [ 0
0
0
1
3
−3
−1
2
−1
−2
−3
3
⋮
⋮
⋮
0
2
−1 𝑣1
−1] [𝑣2 ] = [ 0 ] ⇒ [−1
2 𝑣3
0
−1
−1
2
−1
0
1
0] ⇒ [ 0
0
0
0
0]
0
1
3
0
−2
−3
0
⋮
⋮
⋮
−1
−1
2
⋮
⋮
⋮
0
0]
0
0
0] ⇒ 𝑣1 = 𝑣3 , 𝑣2 = 𝑣3 .
0
𝑣1
𝑠
1
1
For 𝑣3 = 𝑠 ∈ ℝ, 𝒗 = [𝑣2 ] = [ 𝑠 ] = 𝑠 [ 1 ]. The corresponding eigenspace is the plane spanned by [ 1 ].
𝑣3
𝑠
1
1
−1
Therefore, 𝑷 = [ 1
0
−1
0
1
−1/3
1
−1/3
],
whose
inverse
is
[
1
1/3
1
2/3
−1/3
1/3
−1/3
2/3 ]. Thus,
1/3
−1/3
𝑷−𝟏 𝑨𝑷 = [−1/3
1/3
2/3
−1/3
1/3
−1/3
2
2/3 ] [−1
1/3 −1
−1
0
1
1
−1
1] = [−1
1
0
−1
2
−1
−1 −1
−1] [ 1
2
0
2
−1
0
−1 −1
2] [ 1
0
0
−1
0
1
1
3
1] = [0
1
0
0
3
0
0
0]
0
3. Determine whether there exists a 3-by-3 symmetric matrix whose eigenvalues are –1, 3, and 7 and for
0
1
0
which the corresponding eigenvectors are 𝒖 = [ 1 ] , 𝒗 = [0] , and 𝒘 = [1] .
−1
0
1
−𝟏
𝑷𝑫𝑷
0
=[ 1
−1
1
0
0
0 −1
1] [ 0
1
0
0
3
0
0
0
0] [ 1
7 −1
0
= [−1
1
3
0
0
0 0
7] [ 1
7 0
0.5
0
0.5
3
−0.5
0 ]=[0
0
0.5
4 0
4. Let 𝑨 = [2 3
1 0
1
0
0
0 −1
0
1] = [−1
1
1
0
3
4
3
0
0
0
0
7] [ 1
7 −1
1
0
0
0 −1
1]
1
0
4]
3
1
2].
4
a. Find the eigenvalues of 𝑨.
𝜆−4
0
−1
det(𝜆𝑰 − 𝑨) = det [ −2 𝜆 − 3 −2 ] = 0
−1
0
𝜆−4
⇔ (𝜆 − 4)(𝜆 − 4)(𝜆 − 3) − (𝜆 − 3) = 0
⇔ (𝜆 − 3)[𝜆2 − 8𝜆 + 15] = 0
⇔ (𝜆 − 3)(𝜆 − 3)(𝜆 − 5) = 0
⇔ 𝜆1 = 3, 𝜆2 = 3, 𝜆3 = 5
b. For each eigenvalue 𝜆, find the rank of the matrix 𝜆𝑰 − 𝑨.
−1 0 −1
For 𝜆 = 3, 𝜆𝑰 − 𝑨 = [−2 0 −2]. Since the third column of this matrix equals its first column and its
−1 0 −1
second column is the zero vector, 𝜆𝑰 − 𝑨 has only one linearly independent column. Hence, its rank equals
one.
+1 0 −1
For 𝜆 = 5, 𝜆𝑰 − 𝑨 = [−2 2 −2]. Since the second column of this matrix equals −0.5 times the sum of
−1 0 +1
its first and third columns, 𝜆𝑰 − 𝑨 has only two linearly independent columns. Hence, its rank equals two.
c. Is 𝑨 diagonalizable? If so, find 𝑨𝟏𝟎𝟎𝟎 .
0
−1
For 𝜆 = 3, (𝜆𝑰 − 𝑨)𝒙 = 𝟎 is solved for 𝒙 = [1] and 𝒙 = [ 0 ], and their real multiples. For 𝜆 = 5,
0
1
1
(𝜆𝑰 − 𝑨)𝒙 = 𝟎 is solved for 𝒙 = [2], and its real multiples. Then,
1
1 0 −1 5 0 0 1 0 −1 −1
𝑨 = [2 1 0 ] [0 3 0] [2 1 0 ]
1 0 1 0 0 3 1 0 1
and
1 0 −1 5100
1 0 −1 −1
0
0
100
100
𝑨
= [2 1 0 ] [ 0
3
0 ] [2 1 0 ] .
1 0 1
0
0
3100 1 0 1