By using a tweak on the standard basis for $\mathbb F^{n \times n}$, I've shown that the matrix of $T$ with respect to this basis is an $n^2 \times n^2$ upper triangular block matrix, with $n$ blocks of the $n \times n$ $B$ on the diagonal.
Thus, $T$'s characteristic polynomial $c_T(x)$ is simply the product of all the $B$'s characteristic polynomials, i.e. $c_T(x) = [c_B(x)]^n$.
Now, to show that $T$ is diagonalizable if and only if B is diagonalizable, I want to show that when the minimal polynomial $m(x)$ of $T$ or $B$ factors into distinct terms, this implies the other's does also.
But I'm having trouble -- I know theorems (like the minimal polynomial must divide the characteristic), but I can't seem to put these together in a two-way proof. Any help getting traction on this would be much appreciated.
Suppose that $B$ is diagonalizable, let $\{e_1,\ldots,e_n\}$ be a basis s. t. $B(e_i)=c_ie_i$. Consider the basis $e_{ij}$ for ${\mathbb F}^{n\times n}$ s. t. the only non zero coefficient of $e_{ij}$ is the coefficient of the intersection of the $i$-row and $j$-column which is $1$. $T(e_{ij})=c_ie_{ij}$.
Conversely, suppose that $T$ is diagonalizable. There exists a basis $M_j$ of ${\mathbb F}^{n\times n}$ s. t. $BM_j=d_jM_j$. There exist $n$-elements $M_1,\ldots,M_n$ of the basis and vectors $\{u_1,\ldots,u_n\}$ s. t. $\{M_1(u_1),\ldots,M_n(u_n)\}$ is a basis, (if this is not true, this implies that for there exists a vector subspace $V$, $\dim V\lt n$ s. t. $\forall i\in\{1,\ldots,n^2\}\;Im(M_i)\subset V$ and the for every element $\forall M\in\operatorname{span}(M_i),\;Im(M)\subset V$ contradiction.)
$TM_i(u_i)=BM_i(u_i)=d_iM(u_i)\;\implies B$ is diagonalizable in the basis $\{M_1(u_1),\ldots,M_n(u_n)\}$.