# Restriction of compact self adjoint operator to a closed subspace.

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 $T(M)\subset M$ is necessary for the question to make good sense, I think. – paul garrett Nov 28 '16 at 0:01
• @paul garrett can you give a counter example where this doesn't hold. – iron feliks Nov 28 '16 at 0:03
• If $T(v)=\langle v,v_o\rangle\cdot v_o$ for some fixed non-zero $v_o$, then it maps the whole space to $\bb C\cdot v_o$. Unless $M$ contains that complex line, then $T$ restricted to $M$ doesn't even map $M$ to itself. – paul garrett Nov 28 '16 at 0:11

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.