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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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    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).2011-11-01

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The short answer is NO.

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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    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!)2011-11-01