Characteristic polynomial
This is the determinant of
```
 [1–λ  1   0 ]
 [ 0  2–λ  0 ]
 [ 0  –1  2–λ]
```
which is (expand along the first column) (1-λ)(2-λ)².
Eigenvalues
We can read off the roots. They are 1 and 2 (the root 2 has multiplicity 2). These are the eigenvalues.
Bases for eigenspaces
- λ=1
  So
```
 [1–λ  1   0 ]
 [ 0  2–λ  0 ]
 [ 0  –1  2–λ]
```
  becomes
```
 [0  1 0]
 [0  1 0]
 [0 –1 1]
```
  Easily (by inspection or by some row operations) x₁ is free and both x₂ and x₃ must be 0. So a basis for this eigenspace is the set consisting of the single vector
```
[1]
[0]
[0].
```
- λ=2
  So
```
 [1–λ  1   0 ]
 [ 0  2–λ  0 ]
 [ 0  –1  2–λ]
```
  becomes
```
 [–1  1 0]
 [ 0  0 0]
 [ 0 –1 0]
```
  This matrix has rank 2. Both x₁ and x₂ must be 0. A basis for this eigenspace is the set consisting of the single vector (x₃ is free)
```
[0]
[0]
[1].
```

This matrix cannot be diagonalized -- there are not enough linearly independent eigenvectors in the eigenspace corresponding to λ=2.