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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?

Björn Friedrich
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Avishay28
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1 Answers1

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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.

Ben Grossmann
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Kelenner
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  • That was clever :). I was thinking along the lines that det(A)=det(B) but was a dead end :/ –  Jun 23 '17 at 17:02
  • can you kindly tell me is this is a standard technique or did you just think of it? –  Jun 23 '17 at 17:06
  • I have only used the hints given in your question... – Kelenner Jun 23 '17 at 17:10
  • @Kelenner Why P(B) must be invertible? – Avishay28 Jun 23 '17 at 18:06
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    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. – Kelenner Jun 24 '17 at 08:46