Let $H$ be an Hilbert space. Let $F$ be a compact self-adjoint operator on $H$. Show that if $F$ is injective then $H$ is at most countable-dimensional.
I'm not sure how to start this problem…
Below are the definitions I've written down. But I'm not sure how to put the information we have to reach to a conclusion that $H$ is at most countable.
If $F$ is compact, for every bounded sequence $(x_n)$, there exists a subsequence $(x_{n_k})$ such that $F((x_{n_k}))$ converges.
Since $F$ is self-adjoint then $\langle F(x), y\rangle = \langle x, F(y)\rangle$.
Since $F$ is injective, for every $y \in Im(F) $, there exists an unique $y$ such that $F(x)=y$.
Any ideas or hints will be appreciated!
Thanks!
Best Answer
$F(x)=\sum \lambda_n \langle x, e_n \rangle e_n$ for some orthonormal sequence $(e_n)$. If $x$ is orthogonal to each $e_n$ then $F(x)=0$ so $x=0$ by injectivity. Hence $H$ is the closed subspace spanned by $\{e_1,e_2,...\}$.