# Finite dimensional matrices

$A,B$ 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. – copper.hat May 22 '12 at 16:00
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? – Yves May 22 '12 at 16:08
Looks good to me. You can use the Jordan or Schur form to convince yourself of this too. – copper.hat May 22 '12 at 16:09
Thank you, @copper.hat – Yves May 22 '12 at 16:11
@copper.hat You can turn you comment into an answer. – Davide Giraudo May 24 '12 at 15:50

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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Do you mean $B$ and $A^{-1}BA$, or $B$ and $2A^{-1}BA$? – Michael Chen May 24 '12 at 16:23
@MichaelChen: does it matter? – M.B. May 24 '12 at 16:28
Well, if the eigenvalues of $B$ and $A^{-1}BA$ are the same, what does that say about the eigenvalues of $2A^{-1}BA$? – copper.hat May 24 '12 at 16:28
@M.B. Ah, right. Eigenvalues are invariant under scalar multiplication. Thanks. – Michael Chen May 24 '12 at 18:24
@MichaelChen: no, what about I and 2I where I is the identity matrix? – M.B. May 24 '12 at 18:51