# Invariant subspace of 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.

## 1 Answer

Actually you don't need to assume the condition on pure-point spectrum.
The restriction of $T$ to $V$ is a self-adjoint operator from $V$ to $V$, and any self-adjoint operator on a finite-dimensional Hilbert space corresponds to a Hermitian matrix. A Hermitian matrix is diagonalizable.

By the way, a consequence of this is that you can determine the finite-dimensional invariant subspaces: they are the spans of finitely many eigenvectors.

• thank you, so this means that $V^{\perp}$ is an invariant subspace too, right? – MrOperatorTheory Jan 13 '15 at 13:37
• I don't know if it means that, but it's easy to prove. – Robert Israel Jan 13 '15 at 16:02