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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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    @DavideGiraudo: Thanks for the suggestion. Working my way towards$a$T-shirt :-).2012-05-24

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The eigenvalues of $B$ and $A^{-1}BA$ are the same. Conclude something about the eigenvalues. Conclude something about powers of matrices with these eigenvalues.

The Jordan or Schur forms may be useful.

Here's another way of showing that all eigenvalues of $B$ are zero:

Suppose $Bv=\lambda v$, with $\lambda \neq 0$. Then $AB v = \lambda A v = 2 B A v$. From this is follows that $B(Av) = \frac{\lambda}{2} (Av)$. (Since $A$ is invertible, $Av \neq 0$). This shows that if $B$ has a non-zero eigenvalue $\lambda$, then it also has an eigenvalue $\frac{\lambda}{2}$ (hence $\frac{\lambda}{2^k}$, too). This leads to a contradiction since there are at most $n$ distinct eigenvalues. Hence $\lambda = 0$.

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    @Mich$a$elChen: Me too.2012-05-25