Characteristic polynomial
This is the determinant of
```
 [1–λ  1   0 ]
 [ 0  2–λ  0 ]
 [ 0  –1  4–λ]
```
which is (expand along the first column) (1-λ)(2-λ)(4-λ).
Eigenvalues
We can read off the roots. They are 1, 2, and 4. These are the eigenvalues.
Bases for eigenspaces
- λ=1
  So
```
 [1–λ  1   0 ]
 [ 0  2–λ  0 ]
 [ 0  –1  4–λ]
```
  becomes
```
 [0  1 0]
 [0  1 0]
 [0 –1 3]
```
  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  4–λ]
```
  becomes
```
 [–1  1 0]
 [ 0  0 0]
 [ 0 –1 2]
```
  Row operations then produce the matrix
```
 [1 0 –2]
 [0 1 –2]
 [0 0  0]
```
  and a basis for this eigenspace is the set consisting of the single vector (x₃ is free)
```
[2]
[2]
[1].
```
- λ=4
  So
```
 [1–λ  1   0 ]
 [ 0  2–λ  0 ]
 [ 0  –1  4–λ]
```
  becomes
```
 [–3  1 0]
 [ 0 –2 0]
 [ 0 –1 0]
```
  Look at the implied homogeneous system. Both x₁ and x₂ must be 0. A basis for this eigenspace is the set consisting of the single vector
```
[0]
[0]
[1].
```

The matrix can can be diagonalized: A=PDP^–1 where P is

 [1 2 0]
 [0 2 0]
 [0 1 1]

and D is

 [1 0 0]
 [0 2 0]
 [0 0 4].