Let $A,B\in M_n(\mathbb{C})$ such that $A$ has $n$ eigenvalues: $1,2,...,n$, and $B$ also has $n$ eigenvalues: $\sqrt{p_1},\sqrt{p_2},...,\sqrt{p_n}$, in which $p_i$ denotes the $i$-th prime. The question is whether the linear transformation $T\in\operatorname{End}(M_n(\mathbb{C})): X\to AXB$ can be diagonalizable or not.
And in general, if $A$ has $n$ eigenvalues $\{\lambda_i\}_{i=1}^n$, and $B$ has $n$ eigenvalues $\{\mu_i\}_{i=1}^n$, under what restrictions of the eigenvalues we can claim the linear transformation can be diagonalizable?
Now I have a conclusion that: if both of $A,B$ can be diagonalizable, then $T$ above can be diagonalizable, so I wonder if the converse still holds? I.e. if $T$ is diagonalizable, then both of $A,B$ can be diagonalizable, or weaker: one of the $A,B$ can be diagonalizable and $T$ can be diagonalizable, so is the other?
Thanks in advance for any help!
We assume $A\not=O$, $B\not=O$: if either is $O$ then $T=O$ is diagonalisable.
Let $V$ be the $n$-dimensional space. Then we are looking at the map $T=A\otimes B: V\otimes V\to V\otimes V$. Suppose that $T$ is diagonalisable. We prove that both $A$ and $B$ are diagonalisable.
We can decompose $V=\bigoplus U_i$ and $V=\bigoplus W_j$ corresponding to the Jordan-blocks of $A$ and $B$. Then $V\otimes V$ is decomposed into the following $T$-invariant subspaces, $V\otimes V=\bigoplus U_i\otimes W_j$. As $T$ is diagonalisable, it acts on a given $U_i\otimes W_j$ as multiplication by $\lambda_i\mu_j$ where $\lambda_i, \mu_j$ are the eigenvalues of $A,B$ on $U_i, W_j$.
Suppose for a contradiction that some $U_r$ has dimension at least $2$, and has basis $e_1,e_2,\dots$; and that $W_s$ has basis $f_1,\dots$. Then $T:e_2\otimes f_1\mapsto (e_1+\lambda_r e_2)\otimes \mu_s f_1$. This is not multiplication by $\lambda_r\mu_s$ unless $\mu_s=0$. So we are done unless all the $\mu_j=0$.
As $B\not=O$ not all its Jordan blocks are of dimension $1$, so we may swap $A$ and $B$ in the above and again we will be done unless all the $\lambda_i=0$.
Now $A\not=O, B\not=0$ so each has a block of size at least $2$, say $U_1$ with basis $e_1,e_2,\dots$ and $W_1$ with basis $f_1,f_2,\dots$. Then $T:e_2\otimes f_2\mapsto e_1\otimes f_1$, and so $T$ is not multiplication by $0$ after all, and we are done.