---
tag: 0Y95Z7MT02
kind: statement
role: corollary
name: Distinct eigenvalues imply diagonalisable
msc: [15A20]
uses: [056SW1E0AC, 0HNMFG8F8K]
sources:
  - url: https://en.wikipedia.org/wiki/Diagonalizable_matrix
    locator: Characterization
---

## Statement

Let $F$ be a field and let $A$ be an $n \times n$ matrix over $F$.
If $A$ has $n$ pairwise distinct eigenvalues in $F$, then $A$ is [[diagonalisable|07SQCNV233]] over $F$.

## Remarks

Eigenvectors for the $n$ distinct eigenvalues are [[linearly independent|0HNMFG8F8K]], so they form a basis of $F^n$.
The converse fails: $I_n$ is diagonal, but it has only one eigenvalue.
For example, $\begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}$ has eigenvalues $1$ and $3$, so it is diagonalisable.

## Explanations

### L1 human

If an $n \times n$ matrix has $n$ different eigenvalues, then it has $n$ independent eigenvectors.
So it is diagonalisable.
This sufficient condition is easy to check with the characteristic polynomial.