TITLE: Avoiding the Cayley–Hamilton theorem
QUESTION [18 upvotes]: Every $n\times n$ matrix satisfies a polynomial equation of degree at most $n^2$, simply because the space of $n\times n$ matrices has dimension $n^2$.
By the Cayley–Hamilton theorem, every matrix satisfies a polynomial equation of degree $n$.
Is there a simple proof that every matrix satisfies a polynomial equation of degree at most $n$ without using the Cayley–Hamilton theorem?

REPLY [2 votes]: This is a proof by induction:
The result is trivial for $n=1$ and of course we can assume the field $K$ to be algebraically closed.
Pick an eigenvector with $Av = \lambda v$, set $V = \langle v \rangle$. $V$ is an invariant one-dimensional subspace.
We get two new endomorphisms: $A_{|V} \in End(V)$ and $(A_{K^n/V}: [x] \mapsto [Ax]) \in End(K^n/V)$.
By induction we obtain $f$ of degree $1$ with $f(A_{|V})=0$ and $g$ of degree $n-1$ with $g(A_{K^n/V})=0$.
The desired $h(A)=0$ holds for the product $h=fg$:
$g(A_{K^n/V})=0$ translates into $g(A)(K^n) \subset V$, hence $h(A)(K^n)=f(A)(g(A)(K^n)) \subset f(A)(V)=0$.