Let $T: F^{nxn} \to F^{nxn}$ be defined by $T(A) = BA$, where $B$ is some fixed matrix. Find the characteristic polynomial of $T$ (it should have degree $n^{2}$), and prove that $T$ is diagonalizable if and only if $B$ is diagonalizable.
I think I've managed to show (1) that if $v$ is an eigenvector of $B$, then $A = (v|0|...|0)$ is an eigenvector of $T$, and (2) that $\lambda$ is an eigenvalue for $B$ iff $\lambda$ is an eigenvalue for $T$. But how to go about ascertaining the characteristic polynomial and diagonalizability?
Hint: Write the matrix of $T$ with respect to the basis $$\begin{eqnarray*} \{&e_{11},e_{21},\ldots,e_{n1},&\\ &e_{12},e_{22}\ldots,e_{n2},&\\ &\cdots&\\ &e_{1n},e_{2n}\ldots,e_{nn}&\}. \end{eqnarray*}$$ You should find that it is block diagonal.