150 CHAPTER 6. SPECTRAL THEORY
This reduces to (λ− 1)(λ2 − 4λ+ 5
)= 0. The solutions are λ = 1, λ = 2 + i, λ = 2− i.
There is nothing new about finding the eigenvectors for λ = 1 so consider the eigenvalueλ = 2 + i. You need to solve(2 + i)
1 0 0
0 1 0
0 0 1
−
1 0 0
0 2 −1
0 1 2
x
y
z
=
0
0
0
In other words, you must consider the augmented matrix 1 + i 0 0 0
0 i 1 0
0 −1 i 0
for the solution. Divide the top row by (1 + i) and then take −i times the second row andadd to the bottom. This yields 1 0 0 0
0 i 1 0
0 0 0 0
Now multiply the second row by −i to obtain 1 0 0 0
0 1 −i 0
0 0 0 0
Therefore, the eigenvectors are of the form
z
0
i
1
.
You should find the eigenvectors for λ = 2− i. These are
z
0
−i1
.
As usual, if you want to get it right you had better check it. 1 0 0
0 2 −1
0 1 2
0
−i1
=
0
−1− 2i
2− i
= (2− i)
0
−i1
so it worked.
6.2 Some Applications of Eigenvalues and Eigenvec-tors
Recall that n× n matrices can be considered as linear transformations. If F is a 3× 3 realmatrix having positive determinant, it can be shown that F = RU where R is a rotation