Exercise 13, Section 6.4 of Hoffman’s Linear Algebra 
Let $V$ be the space of $n\times n$ matrices over $F$. Let $A$ be a fixed $n \times n$ matrix over $F$. Let $T$ and $U$ be the linear operators on V defined by $T(B)=AB$ and $U(B)=AB-BA$.
(a) True or false? If $A$ is diagonalizable (over $F$), then $T$ is diagonalizable.
(b) True or false? If $A$ is diagonalizable, then $U$ is diagonalizable.

My attempt: (a) Suppose $A$ is diagonalizable. By theorem 6 section 6.4, $A$ is diagonalizable$\iff$$m_A=(x-c_1)\cdots (x-c_k)$, where $c_i\in F$ and $c_i\neq c_j$, if $i\neq j$. By exercise 10 section 6.3, $m_A=m_T$. By theorem 6 section 6.4, $T$ is diagonalizable.
(b) Suppose $A$ is diagonalizable. Let $R:V\to V$ such that $R(B)=BA$. Then $U=T-R$. By (a), $T$ and $R$ are diagonalizable. How to show $T-R$ is diagonalizable?

As a hint towards an alternate, more constructive solution, I'll point out the following: if $v\in F^n$ is an eigenvector of $A$ and $w\in F^n$ is an eigenvector of $A^T$, then the matrix $vw^T\in V$ is an eigenvector of both $T$ and $U$. I'll leave it as an exercise to show that matrices of that form in fact span all of $V$.
