$A,B$ 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.