TITLE: Show that $B$ must equal $0$ if $A$ and $C$ have no common Eigenvalues
QUESTION [6 upvotes]: If $AB=BC$, and $A$ and $C$ have no common eigenvalues, show that $B$ must equal $0$.
How do I do this?

REPLY [3 votes]: Just for grins let's try a Kronecker approach (here is a good source). Expressing $AB-BC=0$ in vector form using Kronecker products gives us
$$\mathop{\textrm{vec}}(AB-BC) = (I\otimes A-C^T\otimes I)\mathop{\textrm{vec}}(B) = 0$$
So if $I\otimes A-C^T\otimes I$ is nonsingular, then $B=0$. Note that 
$$I\otimes A-C^T\otimes I=A\oplus-C^T$$ 
where $\oplus$ denotes the Kronecker sum. If the eigenvalues of $A$ are $\lambda_1,\dots,\lambda_n$ and the eigenvalues of $C$ are $\mu_1,\dots,\mu_n$, the eigenvalues of $A\oplus -C^T$ are 
$$\lambda_i-\mu_j \quad \forall i,j=1,2\dots,n$$
So indeed, it is easy to see that if $A$ and $C$ have distinct eigenvalues, then $I\otimes A-C^T\otimes I$ must nonsingular, because none of the quantities $\lambda_i-\mu_j$ would be zero.
Let's prove that the eigenvalues of $A\oplus -C^T$ are what I say they are. Let $v_i$ be a right eigenvector of $A$ corresponding to $\lambda_i$, and $w_j$ be an left eigenvector of $C$ corresponding to $\mu_j$. Then
$$\begin{aligned}
(I\otimes A-C^T\otimes I)(w_j\otimes v_i)&=w_j\otimes Av_i-C^Tw_j\otimes v_i\\
&=w_j\otimes\lambda_iv_i-\mu_jw_j\otimes v_i\\
&=\lambda_i(w_j\otimes v_i)-\mu_j(w_j\otimes v_i) \\
&=(\lambda_i-\mu_j)(w_j\otimes v_i)\end{aligned}$$
So $w_i\otimes v_i$ is indeed a right eigenvector of our Kronecker sum, with eigenvalue $\lambda_i-\mu_j$.