Show that $B$ must equal $0$ if $A$ and $C$ have no common Eigenvalues If $AB=BC$, and $A$ and $C$ have no common eigenvalues, show that $B$ must equal $0$.
How do I do this?
 A: Sorry for the first comment, it is not so useful to do this by definition.
There is an answer based on $\lambda$ matrix. Let $f_A(\lambda)=|A-\lambda E|$ be the characteristic polynomial of $A$. It is easy to show, as a consquence of $AB=BC$ that $0=f_A(A)B=Bf_A(C)$. What's left is to show that $f_A(C)$ is invertable. 
Let $F_A(\lambda)=\prod_{i=1}^{dim A}(\lambda-\lambda_i)$, then $F_A(C)=\prod_{i=1}^{\dim A}(C-\lambda_iE)$, then $A,C$ has non-comm. eig. implies that $C-\lambda_iE$ must be invertable. If not, then $\exists X\neq0$, such that $CX=\lambda_iX$, which means that $\lambda_i$ is an eig. value of $C$, but it is also an eig. value of $A$, contradiction!
A: Another approach: note that
$$
(A - \lambda I)B = AB  - \lambda B = BC - \lambda B = B(C - \lambda I)
$$
Suppose that $\lambda$ is an eigenvalue of $C$.  It follows, since there are no eigenvalues in common, that $(A - \lambda I)$ is invertible.  Thus, we have
$$
B = (A - \lambda I)^{-1}B(C - \lambda I)
$$
In particular, this is true for every eigenvalue $\lambda$ of $C$.  So, it follows that for the eigenvalues $\lambda_1,\dots,\lambda_n$ of $C$, we have
$$
B = (A - \lambda_1 I)^{-1} B (C - \lambda_1 I) =\\
(A - \lambda_2 I)^{-1} \overbrace{(A - \lambda_1 I)^{-1} B (C - \lambda_1 I)}^B (C - \lambda_2 I) = \\
\vdots\\
(A - \lambda_n I)^{-1} \cdots (A - \lambda_1 I)^{-1} B (C - \lambda_1 I) \cdots (C - \lambda_n I)
$$
Letting $p_C(x)$ be the characteristic polynomial of $C$, this becomes
$$
B = [p_C(A)]^{-1}B[p_C(C)]
$$
by the Cayley-Hamilton theorem, $p_C(C) = 0$, so that the above implies $B = 0$.
A: 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 $\lambda_i-\mu_j$.
A: $\newcommand{\VEig}{\mathrm{VEig}}$
But I also follow my first comment, just go like @AWertheim, we only need to show that if for any $v\in \VEig(C)$ (the eig. vector spance of $C$), we have  $Bv=0$, then $B=0$.
Firstly, if $C$ is diagonalizable, then $\dim\VEig(C)=\dim C$, thus the results holds. But how about $C$ is not diagonalizable?
By the standard theory of LA, we can consider $C$ as its Jordan canonical form, for simplicity, let us suppose $C=\pmatrix{0&1\\0&0}$ be a Jordan block, then you can check that in this case the results holds too.
In fact, $\VEig(C)=\left\{a\pmatrix{1\\0}|a\in R\right\}$, and  $B=\pmatrix{0&b_1\\0&b_2}$. Then $AB=BC=\pmatrix{0&0\\0&0}$. Suppose $A=\pmatrix{\alpha_1\\\alpha_2}$, and let $\beta=\pmatrix{b_1\\b_2}$, then $AB=0$ implies that $\alpha_1,\alpha_2\perp \beta$. If $\beta\neq0$, then $\alpha_1$ and $\alpha_2$ must be linear dependent, which implies that $A$ is of rank $1$ at most, and it must have a non-zero solution of $AX=0$, and $0$ is an eigenvalue of $A$, which is impossible since it is also an eigenvalue of $C$.
For the general case, I suppose it can work in the same way.
