# How does $ABv=BAv$ imply $A^kBv=BA^kv$?

Let $v$ be an eigenvector of $B$ to $\lambda$ which might not be diagonalizable and $ABv=BAv$ with $A$ which might also not be diagonalizable, how does that imply $A^kBv=BA^kv$?

A trivial result that follows is that $Av$ is also an eigenvector of $B$ to the same eigenvalue $\lambda$, so the result would follow if one could show that the associated eigenspace of $B$ is one dimensional which would imply that $v$ is also an eigenvector of $A$

The reason I need this is for a theorem on simultaneous triangularization of matrices.

EDIT: I just realized that if one could show that $A(AB-BA)v=(AB-BA)Av$ then the statement would follow through induction. Can anyone show this?

EDIT2: Can anyone say anything if one assumes that the matrices A and B are diagonalizable?

• you're wrong, that would follow as I already said if one could show that the eigenspace of $B$ to $\lambda$ were onedimensional, what you're saying is that $v$ must also be an eigenvector of $A$ which is not necessarily given. – Not Buying It Sep 3 '13 at 9:21
• "The reason I need this is for a theorem on simultaneous triangularization of matrices." So isn't the hypothesis that $AB = BA$ rather than $AB\nu =BA\nu$, in that case? – Sam Sep 3 '13 at 13:13
• No, that's not the case, the general theorem is stronger, it states that n matrices $A_1, A_2,...A_n$ are triangularizable under an unitary transformation P iff for all polynomials in n noncommuting variables $p(A_1...A_n)[A_i,A_j]$ is nilpotent, which is certainly the case when $A_i,A_j$ commute. – Not Buying It Sep 3 '13 at 13:29

You cannot prove that, because the given conditions do not imply your assertion. Consider $$A=\pmatrix{1&0&0\\ 1&0&0\\ 0&1&2}, \ B=\pmatrix{1&0&1\\ 0&1&1\\ 0&0&2}, \ v=\pmatrix{1\\ 0\\ 0}.$$ Then both $A$ and $B$ are diagonalisable, $Bv=v$ and $ABv=BAv=\pmatrix{1\\ 1\\ 0}$, but $A^2Bv=\pmatrix{1\\ 1\\ 1}\ne\pmatrix{2\\ 2\\ 2}=BA^2v$.
• @Craven The page on your link is unreadable to me (so, probably to the others as well). Unless user1551 made a mistake (which I find highly unlikely), the answer to your question is that there is either an additional condition on $A$, $B$, and/or $v$ which you didn't include in the question, or there is an error in the book. – Vedran Šego Sep 3 '13 at 13:52
• books.google.ch/… does it work now? I've been at it for a while and would like to hit the checkbox, $v$ is stated in this specific example to be an eigenvector of $A_i$ for i from 1 to n-1 and the goal is to conclude with induction that for all given $A_i$ there exists a polynomial $p$ in n variables so that for all elements from the vectorspace $u \in V$ $p(A_1...A_n)u$ is an eigenvector of all the stated matrices – Not Buying It Sep 3 '13 at 13:59
• @Craven The link worked on my computer after I changed the top-level domain from .ch to .com. Anyway, you have left out the condition that "$[AB-BA]f(A)v = 0$ for every polynomial $f$". In the book, the author splits the proof of 40.4 into two cases. In the first case, when $m=2$ (with $A_1=B,\ A_2=A$) and $v$ is an eigenvector of $B$, he assumes that $\color{red}{[AB-BA]f(A)v = 0}$ for every polynomial $f$. So, if we take $f=1$, we get $ABv=BAv$. It follows that $AABv-BAAv=ABAv-BAAv=[A,B]Av$, which is zero because we can take $f(A)=A$. – user1551 Sep 3 '13 at 15:04
If $v$ is an eigenvector of $B$ for eigenvalue $\lambda$ then the condition $ABv=BAv$ implies that $Av$ is also an eigenvector of $B$ corresponding to the same eigenvalue. So $ABAv=BA^2v$. However $ABAv=A^2Bv$ b y the given condition. So we have proved for $k=2$. Now the rest follows by induction.
• How do you get from $ABv = BAv$, $Av$ is a $\lambda$ eigenvector for $B$, to $ABAv = BA^2v$? – Neal Sep 3 '13 at 13:38