I know that the characteristic function of a linear map $T:V\to V$ is defined as $\chi_T(x):=\chi_A(x)$ where $A$ is any matrix for $T$ w.r.t. some basis of $V$. I know this is well-defined as it is independent of choice of basis.
My Question: With $A$ a matrix for $T$, why does it follow that if $\chi_A(A)=0$ then $\chi_T(T)=0$?
I need this for the final step of my proof of Cayley-Hamilton.
A concise way to summarize what's going on here is that, for any fixed basis $B$ of $V$, the map $$T\mapsto[T]_B$$ which sends $T$ to its matrix with respect to $B$ is an isomorphism from the ring of linear transformations of $V$ under composition to the ring of $n\times n$ matrices over the base field $F$ under matrix multiplication, where $n=\dim V$. Therefore if $\varphi(x)\in F[x]$ is any polynomial, $$[\varphi(T)]_B=\varphi([T]_B)$$ Taking $\varphi$ equal to the characteristic polynomial answers your question.