Can I always extend a selfadjoint Operator in $L^2$?

Question

Assume that we have a self-adjoint operator $T\colon D \to D$ where $D \subset L^2$ is some finite dimensional subspace. Can I conclude that than a self-adjoint operator $S \colon L^2 \to L^2$ exists with $S=T$ on $D$ ?

Hope someone can help me. Best regards, Adam

Answer 1

Since $D$ is finite dimensional, it is closed. Let $\Pi$ be the orthogonal projection onto $D$, then let $S = \Pi T \Pi$. (Note that the first $\Pi$ is superflous, but makes it obvious that $S$ is self-adjoint.)

Answer 2

Yes. Take an orthonormal basis $e_1, \dots, e_n$ of $D$, extend it to an orthonormal basis $e_1, e_2, \ldots$ on all of $L^2$ and set $S(e_i) = T(e_i)$ for $i=1, \dots, n$, and $S(e_i) = 0$ for $i>n$.