Let $T$ be a compact and self adjoint operator on a hilbert space $H$ such that $T$ is not invertible. Prove that $\text{Ker}(T)\ne 0$

Question

Let us assume on contrary that $\text{Ker}(T)=0$.

Now we will use a standard result which says $\text{Ker}(T)=\overline{\text{Ran}(T^*)}$

As $T$ is self-adjoint, $T=T^*$. So $\text{Ker}(T)=\overline{\text{Ran}(T)}$.

Therefore, $\overline{\text{Ran}(T)}=H$.

From here, I can't proceed. If I can show $\text{Ran}(T)$ is closed we are done. Although I haven't use the compactness of $T$ yet.

Can anyone help me complete the proof? Thanks for help in advance.

Answer 1

The claim is not true for separable $H$. Define $T:l^2\to l^2$ by $$ Tx = (x_1, x_2/2, x_3/3,\dots), $$ which is compact, self-adjoint, and injective. It is not invertible as $(1,1/2,1/3,\dots)\in l^2$ is not in the range of $T$.

If $H$ is non-separable then the claim follows from the spectral theorem.