Linear Algebra, 2020a

Section II. Similarity 415
We next see the basic tool for finding eigenvectors and eigenvalues.
3.7 Example If
⎛
⎞
1 2 1
⎜
⎟
T = ⎝ 2 0 −2⎠
−1 2 3
then to find the scalars x such that T⃗ζ = x⃗ζ for nonzero eigenvectors ⃗ζ, bring
everything to the left-hand side
⎛
⎞ ⎛
1 2 1
⎜
⎟ ⎜
z ⎞ ⎛
1
⎟ ⎜
z ⎞
1
⎟
⎝ 2 0 −2⎠
⎝z 2 ⎠ − x ⎝z 2 ⎠ = ⃗0
−1 2 3 z 3 z 3
and factor (T − xI)⃗ζ = ⃗0. (Note that it says T − xI. The expression T − x doesn’t
make sense because T is a matrix while x is a scalar.) This homogeneous linear
system
⎛
⎞ ⎛ ⎛ ⎞
⎜
⎝
1 − x 2 1
⎟ ⎜
z ⎞
1
⎟ ⎜
0 ⎟
2 0− x −2 ⎠ ⎝z 2 ⎠ = ⎝0⎠
−1 2 3− x z 3 0
has a nonzero solution ⃗z if and only if the matrix is singular. We can determine
when that happens.
0 = |T − xI|
1 − x 2 1
=
2 0− x −2
∣ −1 2 3− x∣
= x 3 − 4x 2 + 4x
= x(x − 2) 2
The eigenvalues are λ 1 = 0 and λ 2 = 2. To find the associated eigenvectors plug
in each eigenvalue. Plugging in λ 1 = 0 gives
⎛
⎞ ⎛
1 − 0 2 1
⎜
⎟ ⎜
z ⎞ ⎛
1
⎟ ⎜
0
⎞ ⎛
⎟ ⎜
z ⎞ ⎛ ⎞
1 a
⎟ ⎜ ⎟
⎝ 2 0− 0 −2 ⎠ ⎝z 2 ⎠ = ⎝0⎠ =⇒ ⎝z 2 ⎠ = ⎝−a⎠
−1 2 3− 0 z 3 0
z 3 a
for a ≠ 0 (a must be non-0 because eigenvectors are defined to be non-⃗0).
Plugging in λ 2 = 2 gives
⎛
⎞ ⎛
1 − 2 2 1
⎜
⎟ ⎜
z ⎞ ⎛
1
⎟ ⎜
0
⎞ ⎛
⎟ ⎜
z ⎞ ⎛
1
⎟ ⎜
b
⎞
⎟
⎝ 2 0− 2 −2 ⎠ ⎝z 2 ⎠ = ⎝0⎠ =⇒ ⎝z 2 ⎠ = ⎝0⎠
−1 2 3− 2 z 3 0
z 3 b
with b ≠ 0.