Suppose that we have a Hilbert Space $H$ and $M$ is a closed subspace of $H$. Let $T\colon H\rightarrow M$ be the orthogonal projection onto $M$.
I have to show that $T$ is compact iff $M$ is finite dimensional.
So if we assume that $M$ is finite dimensional then $\overline{T(B(0,1))}$ is a closed bounded set in a finite dim vector normed space and so it is compact. Which gives that $T$ is compact.
But I am unsure how to prove that if $T$ is compact then $M$ is finite dimensional?
Thanks for any help
Best Answer
If $M$ is infinite dimensional, then there exists $\{e_n\}_{n\in \mathbb{N}}\subset M$, which is an orthonormal set. Evidently $\{e_n\}_{n\in \mathbb{N}}\subset \overline{T(B(0,1))}$, but $\{e_n\}_{n\in \mathbb{N}}$ has no convergent subsequence, which contraditicts to the compactness of $T$.