Let $H$ a Hilbert space and $f \in H^{'}$. Fixed $z \in H$, $z \neq 0$, define $T: H \rightarrow H$ by $T(x) = f(x)z$. Then $T$ is a compact non-self-adjoint operator. How to determine the spectrum of $ T $?
I verified that $|| T || = || f |||| z || $ and I'm trying to verify that $ \sigma (T) = \{\lambda: | \lambda | \leq || f |||| z || \} $, I have no idea how to make the inclusion $ \sigma (T) \supset \{ \lambda: | \lambda | \leq || f |||| z || \} $. In fact, is it not necessary to place any restrictions on the $ f $ for the result to be true? Or does it work in general?
Since is compact its spectrum consists of eigenvalues, in addition to zero, which is always in the spectrum. To find the nonzero eigenvalues $\lambda$, write $$ f(x)z=\lambda x, $$ which implies that the eigenvector $x$ is a multiple of $z$, and hence can be taken to be $z$ itself, since eigenvalues are scalable. The eigenvalue in turn is computed by solving the equation $$ f(z)z=\lambda z, $$ whose solution is clearly $\lambda=f(z)$. In other words, $$\sigma(T)=\{0,f(z)\}.$$