Show A and B have a common eigenvalue Let A, B and C complex square matrices such that:
$ C\neq 0 $ and $AC=CB $
prove that A and B has a common eigenvalue.
It's worth mentioning that earlier in the assignment I have proved that $A^{n}C=CB^{n}$, 
but I'm not sure how to use it.
This is taken from a linear algebra 2 course.
 A: Let $M$ the minimal polynomial of $A$, and $N$ the minimal polynomial of $B$. We are going to show that $M$ and $N$ have a commun root (and this will prove the assertion). Suppose not. Then there exists $U$, $V$ polynomial such that $M(X)U(X)+N(X)V(X)=1$. Now you have shown that $A^nC=CB^n$ for all $n$. This imply $M(A)C=CM(B)=0$. Hence $CM(B)U(B)=0$ and of course $CN(B)V(B)=0$. We get $C(M(B)U(B)+N(B)V(B))=0=CI=C$, a contradiction. 
A: Let $v_j$ be an eigenvector of $B$ with eigenvalue $\lambda_j$.
Then 
$A C v_j = C B v_j \Rightarrow A (C v_j) = C (\lambda_j v_j) \Rightarrow A (C v_j) = \lambda_j (C v_j) $
Therefore $(C v_j)$ is an eigenvector of $A$ with the same eigenvalue.
A: It has been bugging me to find an answer without (directly) using the minimal polynomial. Here it is:
Since $A^nC=CB^n$, it is easy to see that for any $\lambda$ and $k\ge 0$ we have
$(A-\lambda I)^k C = C (B-\lambda I)^k$.
Suppose $\lambda \not \in \sigma(A)$. Then $C=(A-\lambda I)^{-k} C (B-\lambda I)^k$. In particular, if $v \in \ker (B-\lambda I)^k$, then $Cv=0$.
Hence if the spectra do not overlap, we must have $C=0$ (since $\mathbb{C}^n ={+} _{\lambda \in \sigma(B)} \ker (B-\lambda I)^{m_\lambda} $).
