To clarify, this is a question from a contest--It's from a Chinese contest, a problem from year 2010. $A$ and $B$ are operators $V\to V$ on the complex vector space $V$. I've attempted the problem but seem to have come to some contradictions: First we note $$AB=BA+B$$ And hence if $v$ is an eigenvector of $A$ s.t. $Av=av$, which exists since $A$ is complex, we have $ABv=BAv+Bv=(a+1)Bv$, and hence $Bv$ is also an eigenvector of $A$, with eigenvalue $a+1$.
Proceeding inductively, we have an infinite amount of eigenvalues for $A$, which cannot be possible because each eigenvalue is a root of the characteristic polynomial, which has finite degree, hence $Bv$ has to be zero for each eigen vector of $A$. *not too sure about this part.
Then it remains to show that $A$ has nontrivial kernel,but I'm not sure if my previous arguments have been made correctly.
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LGu
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You are right: if $v \ne 0$ and $Av=av$ then $Bv=0.$ Hence $A$ and $B$ share the common eigenvector $v$. – Fred May 25 '22 at 10:16
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1I’m not convinced that $Bv$ is zero, but only that $B^kv=0$ for some $k$. – Michael Burr May 25 '22 at 10:17
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2Are $A$ and $B$ vectors or matrices? (Likely a typo at the beginning). – Michael Burr May 25 '22 at 10:19
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You’re nearly done. The kernel of $A$ is not needed, if $Bv=0$, then $v$ is an eigenvector for $B$, you don’t need the eigenvalues to be the same. – Michael Burr May 25 '22 at 10:35
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@MichaelBurr Thanks. I did not realize that part! – LGu May 25 '22 at 11:14
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Related: https://math.stackexchange.com/questions/2786438; https://math.stackexchange.com/questions/404412 – user26857 May 25 '22 at 17:25
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If $Bv=0$, then $v$ is an eigenvector of $B$, too.
Suppose $Bv\ne 0$. Inductively $AB^mv=(a+m)B^mv$, so $a+m$ is an eigenvalue of $A$ provided that $B^mv\ne0$. Since $A$ has only finitely many eigenvalues, we must have $B^mv=0$ for some $m\ge2$. Choose $m$ minimal with this property. Then $B^{m-1}v$ is a common eigenvector of $A$ and $B$.
user26857
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