# Are matrices commutable if $ABv=BAv$?

If $A$ and $B$ are both $n \times n$ matrices, and $v$ is a non-zero $n \times 1$ column vector then is it true that if $$ABv = BAv$$ then $$AB=BA$$

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What have you tried? Maybe look at small cases (n = 2 or 3) and see what happens? – Tyler Nov 1 '11 at 2:35
It's not enough to check it for a single random $v$. But if $Xv=Yv$ for all $v$ of the right shape, then $X=Y$. (In particular with $X=AB$ and $Y=BA$). – Henning Makholm Nov 1 '11 at 2:35
If $ABv = BAv$ for $n$ linearly independent $v$, then $AB = BA$. – user17762 Nov 1 '11 at 2:40
No. Here's a counterexample. Let $A = \left(\matrix{2 && 2\\2 && 2}\right), B = \left(\matrix{1 && 5\\ 2 && 4}\right)$ and $v = \left(\matrix{1 \\ 1}\right).$ We have $ABv = BAv = \left(\matrix{24 \\ 24}\right).$ But $AB = \left(\matrix{6 && 18\\6 && 18}\right)$ and $BA = \left(\matrix{12 && 12\\12 && 12}\right).$ – user2468 Nov 1 '11 at 3:10

You cannot say $AB = BA$ if $ABv = BAv$ for some vector $v$.

However, if $ABv = BAv$ is true for all vectors $v$ (or) at-least for $n$ linearly independent vectors $v$, then it is true that $AB = BA$.

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Suppose $AB = C.$ Then we have $ABv= Cv =BAv$ for all $v.$

Do you feel comfortable saying $BA = C$ now? Do you know any theorems about the uniqueness of matrices of linear transformations under fixed bases?

Edit: In Light of the other answer, i should clarify, this is true only for all $v.$

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I have not had very much exposure to linear algebra, and I'm not even sure what "linear transformations under fixed bases" really means. – Olives Nov 1 '11 at 2:39
Well, then I would take the route suggested in the comments above instead of worrying about uniqueness of matrices (this theorem was introduced early in my linear algebra course, so i turn to it i guess!) – barf Nov 1 '11 at 2:42