# Proving that a matrix has a minimal polynomial (without finding it)?

I have been asked to prove that certain square matrix with real entrieshas a minimal polynomial. While finding the actual minimal polynomial is not too hard, I was wondering if there is any way to just prove the existence of the polynomial without going through the steps to find it (just for efficiency sake during an eventual exam)

The fastest way to prove the existence of the minimal polynomial is via some ring theory: If $T$ is a linear operator on a finite-dimensional $F$-vector space $V$, then the set $$I=\{f(x)\in F[x]:f(T)=0\}$$ is an ideal in $F[x]$. Moreover, $I$ is not the zero ideal: if $\dim V=n$, then the vector space of linear maps $V\to V$ has dimension $n^2$, hence the set $\{I,T,T^2,\dots,T^{n^2}\}$ is linearly dependent, which leads to a non-zero polynomial $f$ with $f(T)=0$.
Since $F[x]$ is a principal ideal domain, it follows that $I$ has a unique monic generator, which is the minimal polynomial.