Characteristic polynomial
This is the determinant of
```
 [2–λ 3 ] 
 [ 1 4–λ]
```
which is (2–λ)(4–λ)–3(1)=λ²–6λ+5.
Eigenvalues
The quadratic λ²–6λ+5 factors as (λ–5)(λ–1) so the roots are 5 and 1. These are the eigenvalues.
Bases for eigenspaces
- λ=5
  So
```
 [2–λ 3 ]
 [ 1 4–λ]
```
  becomes
```
 [–3  3 ] 
 [ 1 –1 ]
```
  and the vectors which solve the corresponding homogeneous system all have the difference of their coordinates equal to 0 (the first row is a multiple of the second row, and the second row implies that x₁–x₂=0 so x₁=x₂. So a basis for this eigenspace is the set consisting of the single vector
```
[1]
[1].
```
- λ=1
  So
```
 [2–λ 3 ]
 [ 1 4–λ]
```
  becomes
```
 [1 3] 
 [1 3]
```
  The vectors which solve the corresponding homogeneous system all have x₁+3x₂=0 so that x₁=–3x₂. A basis for that subspace is set consisting of the single vector
```
[–3]
[ 1].
```

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

 [1 –3]
 [1  1]

and D is

 [5 0]
 [0 1].