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$A,B$ are $n\times n$ real matrices and $A$ is non-singular. $AB=2BA$, why then must it follow that $B^n=0$ for some $n$?

Here are some of my thoughts:

We can write $B=2A^{-1}BA$, In other words $B$ is similar to $2B$. So they represent the same transformation but wrt different bases.

Also $B^k=2^kA^{-1}B^kA$. It must have something to do with the finiteness of the matrices

Please just give me a hint. Thank you.

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    The eigenvalues of $B$ and $A^{-1}B A$ are the same. Conclude something about the eigenvalues. Conclude something about powers of matrices with these eigenvalues.2012-05-22
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    Thank you, @copper.hat . So we have that $B$ has the same eigenvalues as $2B$ so all the eigenvalues must be 0 because otherwise if we a non-zero eigenvalue $a$ then $a, 2a, 4a,...$ will also be eigenvalues of $B$ but then $B$ is finite so this cannot be. This kind of matrix must have a power so that it is $0$ because of the characteristic equation. Is this correct?2012-05-22
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    Looks good to me. You can use the Jordan or Schur form to convince yourself of this too.2012-05-22
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    @copper.hat You can turn you comment into an answer.2012-05-24
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    @DavideGiraudo: Thanks for the suggestion. Working my way towards a T-shirt :-).2012-05-24

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