Let $T$ be a compact self adjoint operator on a Hilbert space $H$. Let $M$ $\subset $ $H$ be a closed subspace of $H$. I have to prove $T$ restricted to $M$ is a compact operator.
I'm able to prove this assuming $T(M)$ $\subset$ $M$. However i'm unable to prove that $M$ is invariant under $T$. Any hints?
That $T(M)⊂M $ is not necessary for the question to make good sense !
Let $K$ be the restriction of $T$ to $M$. Then $K:M \to H$ is a linear operator between the Hilbert spaces $M$ and $H$ ( $M$ is closed !)
Now, let $(x_n) $ be a bounded sequence im $M$. Then $(x_n)$ is a bounded sequence in $H$. Since $T$ is compact, $(Tx_n)$ contains a converget subsequence. From $Tx_n=Kx_n$ we derive that $(Kx_n)$ contains a convergent subsequence.
This shows: $K$ is compact.