I'm reading a proof where there's a conclusion: "Since [itex]zW\subset W[/itex], there is an eigenvector [itex]v\neq 0[/itex] of z in W, [itex]zv=\lambda v[/itex]." There W is a subspace of some vector space V, and z is a matrix, in fact a member of some solvable Lie algebra [itex]\mathfrak{g}\subset\mathfrak{gl}(V)[/itex]. (Could be irrelevant information, though.)(adsbygoogle = window.adsbygoogle || []).push({});

This seemed a strange claim, because just because z maps W into W doesn't in general mean that the eigenvector exists like that. However, I noticed that I was unable to come up with a counter example if I allowed the W to be over the complex field. Is there a theorem that says something like this:

Let [itex]W\subset\mathbb{C}^n[/itex] be a vector space, and A a matrix such that [itex]AW\subset W[/itex]. Then there exists an eigenvector [itex]v\in W[/itex] of the matrix A, so that [itex]Av=\lambda v[/itex] for some [itex]\lambda \in\mathbb{C}^n[/itex].

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# Eigenvector existence in complex space

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