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

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?

## marked as duplicate by user1551 linear-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 23 '17 at 17:00

• What if $C=0$ ? – Kelenner Jun 23 '17 at 16:46
• @Avishay28 do you want to recheck the question and edit? – user456218 Jun 23 '17 at 16:47
• I miss that, edited the question.. – Avishay28 Jun 23 '17 at 16:49
• @Avishay28 do we know something about the invertibility of A and B? – user456218 Jun 23 '17 at 16:51
• @AdityaKumar No.. – Avishay28 Jun 23 '17 at 16:54

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.
• 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 8:46