= Solution
View $X$ as the matrix of a <linear map> $T:F^m\to F^n$ of <matrix rank> $r$. Choose vectors $u_1,\ldots,u_r$ whose images form a <basis> of $\operatorname{im}T$, extend them by a basis of $\ker T$ to a basis of $F^m$, and extend $Tu_1,\ldots,Tu_r$ to a basis of $F^n$. In these two bases the matrix of $T$ is the <rank normal form>
$$
\begin{pmatrix}I_r&0\\0&0\end{pmatrix}.
$$
The two changes of basis give invertible $P,Q$ with the displayed matrix equal to $Q^{-1}XP$.
For the <block upper triangular matrix> $A=\begin{pmatrix}B&D\\0&C\end{pmatrix}$, every nonzero term in the <Leibniz formula for determinants> sends the rows belonging to $C$ into the columns belonging to $C$; bijectivity then sends the remaining rows into the $B$ columns. The permutation sum consequently factors into the determinant sums for $B$ and $C$, proving
$$
\boxed{\det A=\det B\det C.}
$$
Finally suppose $L(X)=0$, so $AX=XB$: the map $X:\mathbb C^m\to\mathbb C^n$ is an <intertwining operator>. If $v$ lies in the <generalized eigenspace> of $B$ for $\lambda$, then for some $k$,
$$
(A-\lambda I)^kXv=X(B-\lambda I)^kv=0.
$$
Because $A$ and $B$ have no common <eigenvalue>, $A-\lambda I$ is invertible, hence $Xv=0$. The generalized eigenspaces of $B$ span $\mathbb C^m$, so $X=0$. Thus \b[the <Sylvester equation> operator $L$ is injective.]
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