## The degree of the minimal polynomial of a matrix of dimension n is at most n²

Let $A \in \mathsf{Mat}_n(F)$ be a matrix over a field $F$. Prove that the dimension of the minimal polynomial $m_A(x)$ of $A$ is at most $n^2$.

If $V = F^n$, note that the set $\mathsf{End}_F(V)$ is isomorphic to $\mathsf{Mat}_n(F)$ once we select a pair of bases. The set of all $n \times n$ matrices over $F$ is naturally an $F$-vector space of dimension $n^2$. Consider now the powers of $A$: $1 = A^0$, $A^1$, $A^2$, …, $A^{n^2}$. These matrices, as elements of $\mathsf{End}_F(V)$, are necessarily linearly dependent, so that $\sum r_i A^i = p(A) = 0$ for some $r_i \in F$. The minimal polynomial $m_A(x)$ divides this $p(x)$, and so has degree at most $n^2$.