Suppose we have a Hilbert space $H$ and a compact operator $T$ acting on $H$. If $T$ has no non-zero-eigenvalues, is it necessarily the zero operator?
Secondly, if I decompose $H$ into eigenspaces of $T$, is it true that $T$ is zero on the orthogonal complement of the closure of $E_1+E_2+...$ where $E_n$ denotes the eigenspace of $\lambda_n$ for $\lambda_n$ the non-zero eigenvalues of $T$? I know this must hold if $T$ is Hermitian but not sure about this general case.
No. Any operator on a finite dimensional space is compact and in particular, any non-trivial nilpotent operator $T$ is a counterexample.