Let $A$ and $B$ be $n\times n$ matrices such that $AB=BA$. Show that $A$ and $B$ have a common eigenvector.
I am not able to prove this. Can anyone help?
$\begingroup$ $\endgroup$
12 
- $\begingroup$ BTW A similar statement for Hilbertspaces is the reason for the Heisenberg uncertainty principle in QM, if I remember right. $\endgroup$mvw– mvw2017-02-10 08:54:33 +00:00Commented Feb 10, 2017 at 8:54
- $\begingroup$ What do you mean by "characteristic vector"? $\endgroup$Kal S.– Kal S.2017-02-10 08:57:23 +00:00Commented Feb 10, 2017 at 8:57
- $\begingroup$ Are $A,B$ normal matrices, i.e. does their EVD exist? $\endgroup$GDumphart– GDumphart2017-02-10 09:06:46 +00:00Commented Feb 10, 2017 at 9:06
- $\begingroup$ @Test123 Eigen vectors $\endgroup$Cosmic Focus– Cosmic Focus2017-02-10 09:09:24 +00:00Commented Feb 10, 2017 at 9:09
- $\begingroup$ @user218931 I don't understand the question. Do you mean if A and B have same eigen spaces? $\endgroup$Cosmic Focus– Cosmic Focus2017-02-10 09:39:28 +00:00Commented Feb 10, 2017 at 9:39
| Show 7 more comments
1 Answer
$\begingroup$ $\endgroup$
5 Let $\lambda$ eigenvalue of $B$ and $v\in V_\lambda$ an eigenvector i.e. $Bv=\lambda v$. Then, since $BA=AB$ we have: $$ BAv=ABv=\lambda Av $$ This implies that $Av$ is also an eigenvector of $B$ for the eigenvalue $\lambda$, namely $Av\in V_\lambda$. We deduce that $A V_\lambda \subset V_\lambda$ which is what you are asking as mentioned in the comments (even though the post doesn't clearly mention this).
- $\begingroup$ But how does this prove that Av is also an eigen vector of A? $\endgroup$Cosmic Focus– Cosmic Focus2017-02-13 00:29:14 +00:00Commented Feb 13, 2017 at 0:29
- $\begingroup$ @AnimeshTiwari $Av$ is an eigenvector of $B$, not $A$. $\endgroup$Kal S.– Kal S.2017-02-13 06:40:12 +00:00Commented Feb 13, 2017 at 6:40
- $\begingroup$ yes but one has to prove that A and B have a common eigen vector. Av is an eigen vector of B, but how is it an eigen vector of A as well? $\endgroup$Cosmic Focus– Cosmic Focus2017-02-13 07:23:46 +00:00Commented Feb 13, 2017 at 7:23
- $\begingroup$ @AnimeshTiwari Not true in general. Check the accepted answer here: math.stackexchange.com/questions/1022082/… It's assumed though that $A,B$ are diagonalizable. $\endgroup$Kal S.– Kal S.2017-02-13 07:36:51 +00:00Commented Feb 13, 2017 at 7:36
- $\begingroup$ will go throught that. Thank you! $\endgroup$Cosmic Focus– Cosmic Focus2017-02-13 10:32:15 +00:00Commented Feb 13, 2017 at 10:32