Page 232 - 35Linear Algebra

232 Eigenvalues and Eigenvectors

2 − λ 2
This is a homogeneous system, so it only has solutions when the matrix
16 6 − λ
is singular. In other words,

2 − λ 2
det = 0
16 6 − λ
⇔ (2 − λ)(6 − λ) − 32 = 0

2
⇔ λ − 8λ − 20 = 0
⇔ (λ − 10)(λ + 2) = 0

For any square n × n matrix M, the polynomial in λ given by

n
P M (λ) = det(λI − M) = (−1) det(M − λI)
is called the characteristic polynomial of M, and its roots are the eigenvalues of M.
In this case, we see that L has two eigenvalues, λ 1 = 10 and λ 2 = −2. To find the
eigenvectors, we need to deal with these two cases separately. To do so, we solve the

2 − λ 2 x 0
linear system = with the particular eigenvalue λ plugged
16 6 − λ y 0
in to the matrix.

λ = 10: We solve the linear system

−8 2 x 0
= .
16 −4 y 0

x
Both equations say that y = 4x, so any vector will do. Since we only
4x
need the direction of the eigenvector, we can pick a value for x. Setting x = 1

1
is convenient, and gives the eigenvector v 1 = .
4
λ = −2: We solve the linear system

4 2 x 0
= .
16 8 y 0
Here again both equations agree, because we chose λ to make the system

1
singular. We see that y = −2x works, so we can choose v 2 = .
−2
Our process was the following:

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