256 CHAPTER 12. SPECTRAL THEORY
Here is another more interesting example of a defective matrix.
Example 12.1.14 Let
A =
2 −2 −1−2 −1 −214 25 14
.
Find the eigenvectors and eigenvalues.
In this case the eigenvalues are 3,6,6 where we have listed 6 twice because it is a zeroof algebraic multiplicity two, the characteristic equation being
(λ −3)(λ −6)2 = 0.
It remains to find the eigenvectors for these eigenvalues. First consider the eigenvectors forλ = 3. You must solve
2 −2 −1−2 −1 −214 25 14
−3
1 0 00 1 00 0 1
x
yz
=
000
.
The augmented matrix is −1 −2 −1 | 0−2 −4 −2 | 014 25 11 | 0
and the row reduced echelon form is 1 0 −1 0
0 1 1 00 0 0 0
so the eigenvectors are nonzero vectors of the form(
t −t t)T
= t(
1 −1 1)T
Next consider the eigenvectors for λ = 6. This requires you to solve 2 −2 −1−2 −1 −214 25 14
−6
1 0 00 1 00 0 1
x
yz
=
000
and the augmented matrix for this system of equations is −4 −2 −1 | 0
−2 −7 −2 | 014 25 8 | 0