I'm aware there exist questions (and a wikipedia page) on Sylvester's equation already, but the information there is beyond my current ability (for example I do not know what a kronecker product is or what it means to vectorise). The following question was set in my linear algebra notes, and so I seek a solution using only (elementary) linear algebra.
Let $A\in\mathbb{C}^{n\times n}$, $B\in\mathbb{C}^{m\times m}$ have no common eigenvalue. Show the linear map $\mathbb{C}^{n\times m}\to \mathbb{C}^{n\times m}:X\mapsto AX-XB$ is injective.
We must show $AX=XB\Rightarrow X=0$ and in the case where $B$ is diagonalisable this follows quickly. I cannot solve the general case however. I have tried putting the entries of $X$ into a long vector and finding the matrix representing the linear map but this was too messy. I have also tried choosing a nice basis for, say, $A$ but without being able to simultaneously diagonalise $A$ and $B$ this doesn't seem to help.
Best Answer
Since $A$ and $B$ have no common eigenvalue, if $p$ is the characteristic polynomial of $B$, then $p(A)$ is nonsingular. However, from $AX=XB$, we get $A^kX=XB^k$ for all nonnegative integers $k$. In turn, $p(A)X=Xp(B)=0$. Since $p(A)$ is nonsingular, $X$ must be zero.
Alternatively, by considering two sequence of diagonalisable matrices $\{A_k\}$ and $\{B_k\}$ that converge to $A$ and $B$ respectively, or by considering the matrix representation (which is $I_m\otimes A-B^T\otimes I_n$) of the linear map $L:X\mapsto AX-XB$, it can be shown that the eigenvalues of $L$ are $\lambda_i-\mu_j$, where $\lambda_1,\ldots,\lambda_n$ and $\mu_1,\ldots,\mu_m$ are the eigenvalues of $A$ and $B$ respectively. By assumption, $\lambda_i\ne\mu_j$ for all $i$ and $j$. Hence $L$ is nonsingular.