Trying to prove this fact once you know what you're looking for isn't too hard -- playing around with formulas will show you quickly that if two linear maps commute then they must fix each others' eigenspaces; then just diagonalize the restriction of S to each of T's eigenspaces and stick the resulting mini-bases together.
We can do a little better here: Suppose that we have some commutative collection C of endomorphisms of a finite-dimensional complex vector space V; then they all share some eigenvector.
Proof: We know that C fixes V. Let W be a subspace of V of minimal positive order which is fixed by C. We claim that each vector in W is an eigenvector of each element of C. Suppose not; then there is some counterexample Mw≠λw. M fixes W though, so the restriction of M to W is a complex linear map in its own right; in particular, it has to have some eigenvector in W, say Mx=μx. Let's have Wμ denote the μ-eigenspace of M restricted to W. Then the previous two sentences in effect say that that 0 <>μ <>μ is fixed by C -- take any w in Wμ and any N in C and notice that M(Nw) = (MN)w = (NM)w = N(μw) = μ(Nw), i.e. Nw is a μ-eigenvector of M, and since N, being after all from C, fixes W, it's an element of Wμ. But wait! Before we said that W was as small as you could get and still be fixed by C, but we've just shown that Wμis both fixed by C and smaller than W. This can't be, so our assumption must have been wrong -- W must consist exclusively of eigenvectors for every element of C.Quick question: how many different eigenvalues can be associated with the subspace W in this proof, for any given transformation? For all of them?
Less quick question: What happens when we pass to infinitely many dimensions?
1 comments:
I think is stuff is best done in the context of Lie Algebras. Then there's Lie's theorem, which is true not only for commutative i.e. zero Lie algebras but also for solvable Lie algebras with a very similar proof. Wikipedia calls it Lie-Kolchin theorem. For nilpotent Lie groups, you can do better with Engel's theorem.
As for infinite-dimensional case, you should restrict yourself to some nice class, like compact operators, since for a general case, the question could be very nasty. For compact operators, I think this is true through Lomonosov's theorem.
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