$A,B,C \in M_{n} (\mathbb C)$ and $g(X)\in \mathbb C[x]$ such that $AC=CB$- prove that $A^jC=CB^j$ and $g(A)C=Cg(B)$
Solution 1:
For the eigenvalue part -- Let $g$ be the minimal polynomial of $B$. Since $g(A)C=Cg(B)=0$, if $A$ and $B$ does not share a common eigenvalue, then $g(A)$ is invertible and hence $C=0$, which is a contradiction.
To make $AC=CB$, the matrix $C$ need not be symmetric. Example: $A=\begin{pmatrix}1&0\\0&2\end{pmatrix},\ B=I$ and $C=\begin{pmatrix}0&1\\0&0\end{pmatrix}$.