Let $\mathbb{F}$ be a field,and let $A,B,C$ be matrices over $\mathbb{F}$ of respective sizes $n\times n , k\times k, $and $n\times k$.
put
$M=\begin{bmatrix} A&0 \\ 0&B \end{bmatrix}\quad $ and $\quad N=\begin{bmatrix} A&C \\ 0&B \end{bmatrix}$.
There exists a solution $X\in\mathbb{F}^{n\times k}\quad$ of $\quad AX-XB=C$ $\quad$ if and only if $\quad M$ and $N$ are similar.
- Please give some hints.
Suppose that $M$ and $N$ are similar. Then we may write $MS = SN$ for some invertible matrix $S$. That is, $$ \begin{bmatrix} A&0 \\ 0&B \end{bmatrix}S = S \begin{bmatrix} A&C \\ 0&B \end{bmatrix} $$ Now, we can assume that $S$ has (i.e. that there exists a suitable $S$ with) the block matrix form $\begin{bmatrix}I&X\\0&Z\end{bmatrix}$ (this needs to be justified). Multiplying through on both sides, we have the equality $$ \begin{bmatrix} A & AX \\ 0 & BZ \end{bmatrix} = \begin{bmatrix} A & C + XB \\ 0 & ZB \end{bmatrix} $$ Looking at the top-right entry, we have $AX = C + XB$, which is to say that $AX - XB = C$.
For the converse, it suffices to show that given a particular $X$, the matrix $$ S = \begin{bmatrix} I & X\\0 & I \end{bmatrix} $$ Is an invertible matrix that satisfies $MS = SN$.