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 $p_A(t) = \det(t I_n - A)$. Then $$p_A(A) = 0,$$ where $p_A(A)$ is the value of $p_A$ at $A$.

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 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 $A$.