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This question already has an answer here:

Let $A$, $B$ and $C$ be complex matrices such that $C\neq 0, $ $AC=CB$. Prove that $A$ and $B$ have a common eigenvalue.

There is a hint in the question, these facts can be used for the prove:

  • For a complex matrices $A, B$, If $AB = 0$, and $B$ is invertible, $A = 0$.
  • For a complex matrices $A, B$ and $C$, if $AB = BC$ than for each natural number $k$, $\\\\A^kB = BC^k $.

Any ideas?

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marked as duplicate by user1551 linear-algebra Jun 23 '17 at 17:00

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ What if $C=0 $ ? $\endgroup$ – Kelenner Jun 23 '17 at 16:46
  • $\begingroup$ @Avishay28 do you want to recheck the question and edit? $\endgroup$ – user456218 Jun 23 '17 at 16:47
  • $\begingroup$ I miss that, edited the question.. $\endgroup$ – Avishay28 Jun 23 '17 at 16:49
  • $\begingroup$ @Avishay28 do we know something about the invertibility of A and B? $\endgroup$ – user456218 Jun 23 '17 at 16:51
  • $\begingroup$ @AdityaKumar No.. $\endgroup$ – Avishay28 Jun 23 '17 at 16:54
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By induction, you show that $A^n C =CB^n$ for all $n$. Now take for $P$ the characteristic polynomial of $A$, by the above we deduce that $P(A)C=CP(B)=0$. If no eigenvalue of $B$ is a root of $P$, then $P(B)$ is invertible and $C=0$, contradiction.

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  • $\begingroup$ That was clever :). I was thinking along the lines that det(A)=det(B) but was a dead end :/ $\endgroup$ – user456218 Jun 23 '17 at 17:02
  • $\begingroup$ can you kindly tell me is this is a standard technique or did you just think of it? $\endgroup$ – user456218 Jun 23 '17 at 17:06
  • $\begingroup$ I have only used the hints given in your question... $\endgroup$ – Kelenner Jun 23 '17 at 17:10
  • $\begingroup$ @Kelenner Why P(B) must be invertible? $\endgroup$ – Avishay28 Jun 23 '17 at 18:06
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    $\begingroup$ Let $P(x)=\prod(x-\lambda_k)^{n_k}$ where the $\lambda_k$ are distinct. If none of the $\lambda_k$ are eigenvalue of $B$, Then each of the $B-\lambda_kI$ are invertible, hence $P(B)$ is invertible as product of invertibles matrix. $\endgroup$ – Kelenner Jun 24 '17 at 8:46

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