Page 235 - 35Linear Algebra

12.2 The Eigenvalue–Eigenvector Equation 235

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Example 128 Let L be the linear transformation L: R → R given by
   
x 2x + y − z
y
L   =  x + 2y − z  .
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
L
y 7−→ 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

2
= (λ − 2)[(λ − 2) − 1] + [−(λ − 2) − 1] + [−(λ − 2) − 1]
2
= (λ − 1) (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 −2 −1 0 ∼ 0 −3 −3 0
   
−1 −1 −2 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.

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