$A,B,C \in M_{n} (\mathbb C)$ and $g(X)\in \mathbb C[x]$ such that $AC=CB$- prove that $A^jC=CB^j$ and $g(A)C=Cg(B)$

Let $A,B,C \ne 0 \in M_{n} (\mathbb C)$ and $g(X)\in \mathbb C[X]$ such that $AC=CB$

I need to prove that for every $j=1,2,3..$ the matrices implies $A^jC=CB^j$ and $g(A)C=Cg(B)$ and prove that A and B have a common eigenvalue.

If $AC=CB$ does it mean that $C$ must be diagonal matrix or at list symmetric?

I tried to use Jordan form to solve it, but I didn't succeed.

Thanks again.

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For the common eigenvalue part: Are you sure you don't have to require that $C\not=0$? – Nick Strehlke Aug 14 '11 at 21:53
Sorry, you are right. – user6163 Aug 14 '11 at 21:57
You can prove the first part by simple induction. – Jozef Aug 23 '11 at 13:40

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 $C=\begin{pmatrix}0&1\\0&0\end{pmatrix}$.

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Very nice argument :) – Beni Bogosel Aug 15 '11 at 8:29
The advantage here is that you do not require the ground field to be algebraically closed, which my argument does (by assuming the existence of a Jordan canonical form). Nice. – Arturo Magidin Aug 15 '11 at 16:31
Dear @Arturo: I'm afraid I don't understand you comment. I would have said that the assumption that the ground field is algebraically closed is used in the answer, in the form: two polynomials having no common root are relatively prime. – Pierre-Yves Gaillard Aug 23 '11 at 16:10
@Pierre-Yves: You're right, of course. What I was thinking is that if the minimal polynomials of $A$ and $B$ are relatively prime, then the argument holds, but $A$ and $B$ can fail to have common eigenvalues over, say, $\mathbb{R}$, and yet have non-relatively prime minimal polynomials. – Arturo Magidin Aug 23 '11 at 16:26

For the eigenvalues part: If $\lambda$ is eigenvalue of $B$ so there's $v$ such that $Bv=\lambda v$

$ACv=CBv=C\lambda v= \lambda Cv$ so $Cv$ is an eigenvector for $\lambda$ so they share $\lambda$..

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not if $Cv=0$, be careful! – Henri Aug 23 '11 at 14:02
but $C \neq 0$, and v as well. Thanks for correction! – Jozef Aug 23 '11 at 14:21
$C$ is not invertible a priori, so you could have $Cv=0$ and $v\neq 0$. – Henri Aug 23 '11 at 14:48