if Y is diagonalizable, show all eigenvaues are of the form… Given: $X\in\mathcal{M}_n(\mathbb{Q})$ and $Y\in\mathcal{M}_m(\mathbb{Q})$
$R : \mathcal{M}_n (\mathbb{Q}) \to \mathcal{M}_m(\mathbb{Q})$
mapping from matrix $n \times n$ to $m \times m$
$R(Z) = XZ-ZY$
Show:
if $Y$ is diagonalizable then all eingenvalues of $R$ are of the form $p-q$ with $p$ (resp. $q$) an eingenvalue of $X$ (resp. $Y$).

My thought: I have no clue how to even begin.
 A: Let $P$ be a matrix which diagonalizes $Y$. Then we may write
$$Y = PDP^{-1}$$
Now let us transform $R$ according to the above
$$R(Z) = XZ - ZY = XZ - ZPDP^{-1} = (XZP - ZPD)P^{-1}$$
If we define the associated linear map $R_D$ as
$$R_D(Z) = ZX - ZD$$
what the above says is that 
$$R_D(ZP) = R(Z)P$$
Now notice that if $Z$ is an eigenvector of $R$ under eigenvalue $\lambda$, then we have the associated eigenvector $ZP$ for $R_D$ with the same eigenvalue:
$$R(Z) = \lambda Z \iff R_D(ZP)= R(Z)P=\lambda ZP$$
Therefore it suffices to examine the eigenvalues of $R_D$. Now let $Z$ be an eigenvector of $R_D$ with eigenvalue $\lambda$. We then have
$$R_D(Z) = XZ - ZD = \lambda Z \iff XZ = Z(\lambda I + D)$$
This automatically implies that every (non-zero) column of $Z$ must be an eigenvector for $X$. If $\mathbf{z}_i$ denotes the $i$th column of $Z$ and $d_i$ denotes the $i$th entry of $D$ (i.e. the $i$th eigenvalue of $Y$), then
$$X\mathbf{z}_i = (\lambda + d_i)\mathbf{z}_i$$
Therefore each $\lambda + d_i$ is an eigenvalue of $X$ and in particular $\lambda$ is equal to $p - q$ where $p$ is an eigenvalue of $X$ and $q$ is an eigenvalue of $Y$.
