About Linear algebra question. Let $F$ be a field, $n$ a positive integer, and $V$ be the space of $n \times n$ matrices over $F$. If $A$ is a fixed $n \times n$ matrix over F,  $ T_A(B) = AB - BA$. Consider the family of linear operators $T_A$ obtained by letting $A$ vary over all diagonal matrices. 
Prove that the operators in that family are simultaneously diagonalizable.
 A: Hint:
Prove that $T_{A_1}$ and $T_{A_2}$ commute  for all  diagonal  $A_1,A_2$.
A: Alternatively, you may prove that for every diagonal matrix $A$, the matrix representation of $T_A$ w.r.t. the standard basis of $M_n(\mathbb{F})$ is already diagonal. (You may need Kronecker product and the vec operator.) So, there is no need to diagonalise it in the first place.
A: Hints for you to understand and prove:
1) Suppose $\,A,B\,$ are simultaneously diagonalizable (=s.d.) , then there exists invertible $\,P\,$ s.t.:
$$D_1=P^{-1}AP\;,\;\;D_2=P^{-1}BP\;\text{are diagonal}\implies A=PD_1P^{-1}\;,\;B=PD_2P^{-1}\;,\;$$
$$AB=PD_1P^{-1}PD_2P^{-1}=PD_1D_2P^{-1}=PD_2D_1P^{-1}=PD_2P^{-1}PD_1P^{-1}=BA$$
2) Suppose $\,A,B\,$ commute and both have minimal polynomials without multiple roots (iff the min. pol. is a product of different linear factors iff the matrix/operator is diagonalizable). Let $\,\lambda\,$ be an eigenvalue of $\,A\,$ and let $\,V_{A,\lambda}\,$ be the corresponding eigenspace, so:
$$v\in V_{A,\lambda}\implies A\color{red}{Bv}=BAv=B(\lambda v)=\lambda\color{red}{Bv}\implies Bv\in V_{A,\lambda}\implies$$
$\,V_{A,\lambda}\,$ is $\,B$- invariant, so if $\,\beta\,$ is a basis of the linear space formed by eigenvectors of $\,A\,$, the matrix representation of $\,B\,$ wrt $\,\beta\,$ is a block one, with one block for each different eigenvalue of $\,A\,$ , but
(i) A block matrix is diagonalizable iff each block is
(ii) There exists a new basis of $\,V_{A,\lambda}\,$ under which B is diagonal... but also $\,A\,$ is (why?) .
Complete the exercise.
