# Common eigenvector of linear operator

Let $T$ and $S$ be two symmetric, compact linear operators on a (separable) Hilbert space $H$ that commute. Why is there at least 1 common eigenvector of $T$ and $S$?

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Doesn't the usual proof for commuting transformations work? If $\lambda$ is an eigenvalue of $T$ and $W$ is the eigenspace corresponding to $\lambda$, then for every $w\in W$ $T(Sw) = S(Tw) = \lambda(Sw)$, so $S(W)\subseteq W$; restrict $S$ to $W$ and find an eigenvector for $S|_W$. (Note the common eigenvector may corespond to different eigenvalues). –  Arturo Magidin Apr 16 '12 at 1:50
@Arturo: of course it does, provided you know that there are eigenvectors for either $S$ and $T$; this is guaranteed by (an easy part of) the spectral theorem for symmetric (= self-adjoint) compact operators. –  t.b. Apr 16 '12 at 2:00
@t.b.: Well, right; same reason you would want "symmetric" in the finite-dimensional case... –  Arturo Magidin Apr 16 '12 at 2:02

Two commuting compact operators are simultaneously diagonalizable. This means that there exists an orthonormal basis such that both $S$ and $T$ are diagonal with respect to that basis, i.e. every vector in such a basis is a common eigenvector.