The question:
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 can only prove the converse. Assume $t$ is a characteristic value of $T$, then $\det(tI-T)=0$. To show that $c=f(t)$ is a characteristic value for $f(T)$, we let $f(x)=b_0+b_1x+\cdots+b_nx^n$ and observe that \begin{align*} \det(cI-f(T))&=\det(f(t)I-f(T))\\ &=\det(b_1(tI-T)+b_2(t^2I-T^2)+\cdots+b_n(t^nI-T^n))\\ &=\det(tI-T)\det(b_1I+b_2(tI+T)+\cdots+b_n(t^{n-1}I+\cdots+T^{n-1}))\\ &=0. \end{align*} This proved the converse. But I cannot guarentee that the longer one in the determinant is nonzero, if I can ensure this I could prove the other side, also note that I haven't use the information about algebraically closeness of $F$, I think this is crucial, but I don't know how to use it. Any suggestion?
For example $T$ is similar to an upper triangle matrix, i.e. $T=PJP^{-1}$, where $J$ is upper triangle with diagonal entries $t$ running through the eigenvalues of $T$. Thus $f(T)=P^{-1}f(J)P$, and $f(J)$ is an upper triangle matrix with diagonal entries $f(t)$.