TITLE: A compact operator on a non-separable Hilbert space has $0$ in its point spectrum
QUESTION [5 upvotes]: If $H$ is a non-separable Hilbert space and $T:H \to H$ be compact , then show that $0 \in \sigma_{p}(T)$. 
I have no idea about working on a non-separable Hilbert space. It is obvious that it won't have a countable orthonormal basis. Apart from that I don't know anything about non-separble Hilbert Space. Thanks in advance !!!

REPLY [7 votes]: The point is to show that $\ker T\ne\{0\}$. 
Assume first that $T$ is selfadjoint. 
Since $T$ is compact, every nonzero element of the spectrum is an eigenvalue with finite multiplicity, and there are at most countably many nonzero eigenvalues. That is, $$ T=\sum_{k=1}^\infty\lambda_kP_k,$$ where $\{P_j\}$ are pairwise orthogonal finite-rank projections. 
As $Tx=\sum_{k=1}^\infty\lambda_kP_kx$, we deduce that the image of $T$ is (at most) countably-dimensional. Now, since 
$$
H=\ker T\oplus \overline{\text{ran}\,T},
$$
it follows that $\ker T$ is nonzero; actually, it has uncountable dimension. 
For non-selfadjoint $T$, use the fact that $\ker T=\ker T^*T$.