Let $T$ be a linear operator on a finite dimensional vector space over an algebraically closed field $F.$ Let $f$ be a polynomial over $F.$ Prove that $c$ is a characteristic value of $f(T)$ iff $c=f(t)$, where $t$ is a characteristic value of $T.$
I have proved that $c$ is a characteristic value of $f(T)$, where $c=f(t)$ and $t$ is a characteristic value of $T$, but I cannot show rigorously that all characteristic values of $f(T)$ are of the form $f(t)$ where $t$ is the characteristic value of $T$.By intuition I know it is so $K$ is an algebraically closed field and because if the space $V$ has dimension $n$,then $T$ has $n$ eigenvalues (counted with multiplicity) and $f(T)$ also has also $n$ eigenvalues (counted with multiplicity). So, since for all $n$ eigenvalues $t$ of $T$, I have $n$ eigenvalues $f(t)$ of $f(T)$ then I exhaust the eigenvalues of $f(T)$ but I feel that this way of proving is not rigourous.