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I'm preparing for my final and this question came up in one of the practices. I am tempted to say no, but I've been having trouble proving this.

If a vector $v$ is an eigenvector of both matrices $A$ and $B$, is $v$ necessarily an eigenvector of $AB$?

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Hint: Matrix multiplication is associative: $$AB(v)= A(Bv)$$

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  • $\begingroup$ I love hints like this. Thank you! $\endgroup$ – Joseph Jul 20 '13 at 16:06
  • $\begingroup$ @Joseph: No problem! I like giving hints like this too. Much less typing ;) $\endgroup$ – Sharkos Jul 20 '13 at 16:09
  • $\begingroup$ Have you any idea whether associativity is really the crucial property? I'm rather fascinated by translating this from the trivial $\mathbb{R}^{n\times n}$ or $\mathbb{C}^{n\times n}$ cases to the quaternions (where the statement in question seems to hold true) or octonions (no idea). $\endgroup$ – leftaroundabout Jul 21 '13 at 17:20
  • $\begingroup$ Urgh, octonion matrices?! Well, I'd certainly imagine you'd find the non-associativity to be inherited by the matrix multiplication, and then you should be able to construct a 2x2 counterexample... Just make both matrices diagonal and choose the components so that they scale differently from each other, perhaps? $\endgroup$ – Sharkos Jul 21 '13 at 19:43
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If $v$ is an eigenvector associated to the eigenvalue $\lambda$ of $A$ and $\mu$ of $B$ then $$ABv=A(Bv)=A(\mu v)=\mu Av=\mu\lambda v$$ so $v$ is an eigenvector associated to the eigenvalue $\mu\lambda$ of $AB$.

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  • $\begingroup$ What if the entries are from a strictly skew field, though? $A(\mu v)=\mu A v$ isn't valid then, yet the statement appears to hold also for quaternion matrices. $\endgroup$ – leftaroundabout Jul 20 '13 at 19:25
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Note that $(AB)v=A(Bv)=A(\beta v)= \beta(Av)=(\beta\alpha ) v$ with $\alpha$ and $\beta$ being the relevant eigenvalues.

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