Lecture Notes in Differential Equations - Bruce E. Shapiro

124 LESSON 14. REVIEW OF LINEAR ALGEBRA
Its characteristic equation is
2 − λ −2 3
0 =
1 1 − λ 1
∣ 1 3 −1 − λ ∣
(14.44)
= (2 − λ)[(1 − λ)(−1 − λ) − 3] + 2[(−1 − λ) − 1] (14.45)
+ 3[3 − (1 − λ)] (14.46)
= (2 − λ)(−1 + λ 2 − 3) + 2(−2 − λ) + 3(2 + λ) (14.47)
= (2 − λ)(λ 2 − 4) − 2(λ + 2) + 3(λ + 2) (14.48)
= (2 − λ)(λ + 2)(λ − 2) + (λ + 2) (14.49)
= (λ + 2)[(2 − λ)(λ − 2) + 1] (14.50)
= (λ + 2)(−λ 2 + 4λ − 3) (14.51)
= −(λ + 2)(λ 2 − 4λ + 3) (14.52)
= −(λ + 2)(λ − 3)(λ − 1) (14.53)
Therefore the eigenvalues are -2, 3, 1. To find the eigenvector corresponding
to -2 we would solve the system of
⎛
⎞ ⎛ ⎞ ⎛ ⎞
2 −2 3 x x
⎝1 1 1 ⎠ ⎝y⎠ = −2 ⎝y⎠ (14.54)
1 3 −1 z z
for x, y, z. One way to do this is to multiply out the matrix on the left and
solve the system of three equations in three unknowns:
2x − 2y + 3z = −2x (14.55)
x + y + z = −2y (14.56)
x + 3y − z = −2z (14.57)
However, we should observe that the eigenvector is never unique. For example,
if v is an eigenvector of A with eigenvalue λ then
A(kv) = kAv = kλv (14.58)
i.e., kv is also an eigenvalue of A. So the problem is simplified: we can
try to fix one of the elements of the eigenvalue. Say we try to find an
eigenvector of A corresponding to λ = −2 with y = 1. Then we solve the
system
2x − 2 + 3z = −2x (14.59)
x + 1 + z = −2 (14.60)
x + 3 − z = −2z (14.61)