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?

  • $\begingroup$ 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. $\endgroup$ – Not Buying It Sep 3 '13 at 9:21
  • $\begingroup$ "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? $\endgroup$ – Sam Sep 3 '13 at 13:13
  • $\begingroup$ 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. $\endgroup$ – 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 $A^2Bv=\pmatrix{1\\ 1\\ 1}\ne\pmatrix{2\\ 2\\ 2}=BA^2v$.

| cite | improve this answer | |
  • $\begingroup$ then could you please explain why it says in here under mark 1) said thing? Did I read the requirements wrongly? And thanks for finding the example $\endgroup$ – Not Buying It Sep 3 '13 at 13:35
  • $\begingroup$ @Craven I have no access to the paragraph you are talking about. Apparently, the linked page on Google Books is not available to all computers. $\endgroup$ – user1551 Sep 3 '13 at 13:50
  • $\begingroup$ @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. $\endgroup$ – Vedran Šego Sep 3 '13 at 13:52
  • $\begingroup$ 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 $\endgroup$ – Not Buying It Sep 3 '13 at 13:59
  • 2
    $\begingroup$ @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 $f(A)=A$. $\endgroup$ – 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.

| cite | improve this answer | |
  • $\begingroup$ How do you get from $ABv = BAv$, $Av$ is a $\lambda$ eigenvector for $B$, to $ABAv = BA^2v$? $\endgroup$ – Neal Sep 3 '13 at 13:38
  • $\begingroup$ Sorry for the terrible mistake. @Neal is correct. $\endgroup$ – Abishanka Saha Sep 3 '13 at 13:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.