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I'm trying to prove the following:

If $S\colon V\to V$ and $T\colon V\to V$ are unitary linear transformations on unitary space $V$ ($\dim V=n$, $n$ is finite), such that $ST=TS$, then they have a joint eigenvector basis (aka there is a basis of $V$ composed of eigenvectors of both $S$ and $T$ - not necessarily of the same eigenvalue per each).

Can anyone help me out? I've tried rephrasing the 'matrix equivalent' of the theorem, but I didn't get much further.

Thanks!

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    Hint: Since $S$ and $T$ commute, show that each eigenspace of $S$ is $T$-invariant (that is, if $Sv = \lambda v$, we have $S(Tv) = \lambda (Tv)$.2011-08-12
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    This can be strengthened to the conclusion that the basis is orthonormal. @iroiroaru: Are you able and willing to use the fact that a single unitary transformation has an eigenvector (orthonormal) basis?2011-08-12
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    Hi Geoff, I've actually realized that but I wasn't able to see how it "helps me out"... I guess I just have no idea how to start building the actual basis. Jonas: yes, certainly, we've covered it in class.2011-08-12
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    @iroiroaru: It's difficult for me to go much further without telling you the whole answer. As Jonas says, you really need to use the fact that a single unitary linear transformation has an othonormal basis of eigenvectors.2011-08-12
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    BTW there is a really beautiful abstract proof of a somewhat more general statement here: http://planetmath.org/encyclopedia/CommutingMatrices.html2011-08-13

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