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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

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3 Answers 3

up vote 1 down vote accepted

Assume $T$ compact. As $T(M\cap B(0,1))=M\cap B(0,1)$, then $M\cap B(0,1)$ has a compact closure. Conclude by Riesz theorem (which is easier to prove in the context of Hilbert spaces).

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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 $T$.

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My first attempt would have been what the answer by Davide and Richard did. But here's another approach.

Since $T$ is an orthogonal projection, it is a positive operator. The equation $T^2=T$ guarantees that all eigenvalues are either $1$ or $0$. As $T$ is compact, the multiplicity of $1$ as an eigenvalue is finite (because $0$ is the only possible accumulation point in the spectrum of a compact operator). So $\dim\, M=\text{Tr}\,(T)<\infty$.

(depending on context, the last inequality might not be obvious. Then we could just say that $T$ is a finite sum of rank-one projections, i.e. it is a finite-rank projection. So $M$, being the range of a finite-rank projection, is finite-dimensional)

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