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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.

2 Answers2

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

user1551
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    Very nice argument :) – Beni Bogosel Aug 15 '11 at 08:29
  • 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
  • @user1551 why g(A) must be invertible? – Avishay28 Jun 23 '17 at 19:56
  • @Avishay28 $A$ and $B$ do not share any eigenvalues. So, $g(a)\ne0$ for every eigenvalue of $A$. Hence $g(A)$ is invertible. Alternatively, if $A$ and $B$ do not share a common eigenvalue, then their minimal polynomials --- say $f$ and $g$ --- are relatively prime. So, there exist polynomials $p,q$ such that $pf+qg=1$. Since $f(A)=0$, we get $q(A)g(A)=I$. Hence $g(A)$ is invertible. – user1551 Jun 23 '17 at 20:58
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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 $\lambda$..

Jozef
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