linear-guest

12.2 The Eigenvalue–Eigenvector Equation 235
Example 128 Let L be the linear transformation L: R 3 → R 3 given by
⎛ ⎞
x
⎛
2x + y − z
⎞
L ⎝y⎠ = ⎝
z
x + 2y − z
−x − y + 2z
⎠ .
In the standard basis the matrix M representing L has columns Le i for each i, so:
⎛ ⎞ ⎛
⎞ ⎛ ⎞
x 2 1 −1 x
⎝y⎠
↦−→
L ⎝ 1 2 −1⎠
⎝y⎠ .
z −1 −1 2 z
Then the characteristic polynomial of L is 2
⎛
⎞
λ − 2 −1 1
P M (λ) = det ⎝ −1 λ − 2 1 ⎠
1 1 λ − 2
= (λ − 2)[(λ − 2) 2 − 1] + [−(λ − 2) − 1] + [−(λ − 2) − 1]
= (λ − 1) 2 (λ − 4) .
So L has eigenvalues λ 1 = 1 (with multiplicity 2), and λ 2 = 4 (with multiplicity 1).
To find the eigenvectors associated to each eigenvalue, we solve the homogeneous
system (M − λ i I)X = 0 for each i.
λ = 4: We set up the augmented matrix for the linear system:
⎛
−2 1 −1
⎞
0
⎝ 1 −2 −1 0⎠
−1 −1 −2 0
∼
∼
⎛
1 −2 −1
⎞
0
⎝0 −3 −3 0⎠
0 −3 −3 0
⎛ ⎞
1 0 1 0
⎝0 1 1 0⎠ .
0 0 0 0
⎛ ⎞
−1
Any vector of the form t ⎝−1⎠ is then an eigenvector with eigenvalue 4; thus L
1
leaves a line through the origin invariant.
2 It is often easier (and equivalent) to solve det(M − λI) = 0.
235