---
tag: 0F765KFC2J
kind: statement
role: theorem
name: Cayley–Hamilton theorem
msc: [15A24, 15A18]
uses: [039HF58KQ5, 0BNEXSKNRV]
sources:
  - url: https://en.wikipedia.org/wiki/Cayley–Hamilton_theorem
    locator: lead section
same_as:
  wikidata: Q656772
  wikipedia: Cayley–Hamilton theorem
---

## Statement

Let $K$ be a commutative ring, let $n \ge 1$, and let $A$ be an $n \times n$ matrix over $K$ with [[characteristic polynomial|0VSSPAFM2C]] $p_A(t) = \det(t I_n - A)$.
Then
$$p_A(A) = 0,$$
where $p_A(A)$ is the [[value of $p_A$ at $A$|0E3MV6E6MG]].

## Remarks

For $n = 2$, it says $A^2 - (\tr A) A + (\det A) I_2 = 0$.
Substituting $t = A$ into $\det(t I_n - A)$ is not a valid proof, because $p_A(A)$ is a matrix and $\det(A - A)$ is a scalar.
One proof uses the [[adjugate|07W9H8QF38]] identity $(t I_n - A) \operatorname{adj}(t I_n - A) = p_A(t) I_n$.
If $A$ is invertible, then the theorem expresses $A^{-1}$ as a polynomial in $A$.

## Explanations

### L1 human

Every square matrix satisfies its own characteristic polynomial.
For $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$, the polynomial is $t^2 - 5t - 2$, so $A^2 = 5A + 2I$.
So every power of $A$ is a combination of $I$ and $A$.