This post proves Cayley-Hamilton using invariant subspaces.

The Cayley-Hamilton theorem is one of the most basic results in linear algebra.

(Cayley-Hamilton) If $A$ is a $n \times n$ matrix with entries in a field $F$, $p_A(t) = det(tI_n - A)$ is its characteristic polynomial, then $p_A(A) = 0$.

Proof

To show that $p(A) = 0$, it suffices to show that for any vector $v \in F^n$, $p(A)v = 0$. This prompts us to think about invariant subspaces.

Take any $v \neq 0 \in F^n$. Let $W = \mathrm{span} \{ v, Av, \cdots \}$ be a subspace of $F^n$. If $\mathrm{dim} W = k$, it means that the minimal polynomial of $v$ is of degree $k$.  Let this polynomial be $q(t) = a_0 + a_1t + \cdots + a_{k}t^{k}$.

Extend a basis of $W$, namely $v, Av, \cdots, A^{k-1}v$ to a basis of $V$. Then the upper $k \times k$ block $B$ of the matrix wrt the new basis is exactly $\left(\begin{array}{ccccc} 0 & 0 & \cdots & 0 & -a_0 \\ 1 & 0 & \cdots & 0 & -a_1 \\ 0 & 1 & \cdots & 0 & -a_2 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & \cdots & 1 & -a_n \end{array}\right)$

Since the characteristic polynomial is an invariant under base change, and by the product formula of determinant of block matrices, we have $p_B(t) | p_A(t)$. It remains to show that $p_B(t) = q(t) = a_0 + a_1t + \cdots + a_{k}t^{k}$, which is easily checked by direct computation.

Since $q(A)v = 0$, we have $p_A(A)v = 0$. Since $v$ is arbitrary, we have shown that $p_A(A) = 0$.

Remark

1. The above proof shows that Cayley-Hamilton holds in any field $F$. According to the wikipedia entry, this theorem is in fact the source of Nakayama’s lemma.