# Invariant subspace of bounded self-adjoint operator

Assume that we have a self-adjoint operator $T: H \rightarrow H$ with a representation $T(x) = \sum_{n=0}^{\infty} \lambda_n \langle x , x_n \rangle x_n,$ so we have pure point spectrum.

Now, I was wondering whether this classical Linear Algebra result still holds:

Let $V \subset H$ be a finite-dimensional subspace such that $T(V) \subset V$, then $T|_V$ is a diagonizable matrix?

Somehow, this sounds very natural, since $V$ is closed, so we can decompose the operator in $T: V \oplus V^{\perp} \rightarrow H$, but I have difficulties to characterize the invariant spaces.

## migrated from mathoverflow.netJan 13 '15 at 17:49

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• It is enough to observe that $T$ is still symmetric on $V$. – user138530 Jan 13 '15 at 1:04
• Is $(x_n)$ an orthonormal basis for $H$? – Janko Bracic Jan 13 '15 at 20:30