Given $\psi$ and $\phi$ in a Hilbert space $H$, we let $T$ be the rank-1 operator such that $$T\varphi=<\psi,\varphi>\phi.$$ It is easy to find the eigenvalues of $T$, they are $0$ and $<\psi,\phi>$. It is a bit harder to find the spectrum of $T$, as well as the point spectrum, continuous spectrum, and remainder spectrum of $T$.
The point spectrum (i.e., the set of $x\in\mathbb{C}$ such that $T-x$ is not injective) is pretty straightforward, it is simply the eigenvalues by looking at the kernel of $T$.
The eigenvalues are also in the spectrum, but I'm having a hard time determining if there are any other points in it.
The continuous spectrum (i.e. the set of $x\in\mathbb{C}$ such that $T-x$ is injective and the image of $T-x$ is dense) definitely does not include the eigenvalues. I tried thinking about a sequence $f_n$ that converges to every $f$ in $H$, but I couldn't come up with anything. My gut feeling is that $T-x$ is dense for every $x$ except the eigenvalues, as the image of $T$ is the span of $\phi$ in $H$ and I think this is dense.
My argument above would make the remainder spectrum (i.e. the set of $x\in\mathbb{C}$ such that $T-x$ is injective and the image of $T-x$ is not dense) the empty set.
Any help would be appreciated.


Any finite rank operator is a compact operator, and it's a known result that the only points in the spectrum of a compact operator are the eigenvalues. See theorem 35.17 here for the general statement.

  • $\begingroup$ Great, thank you for the reply! So since the spectrum consists of the eigenvalues and the point spectrum consists of the eigenvalues as well, does this mean the continuous spectrum and residual spectrum are empty? I've read online that the spectrum is the disjoint union of the three sets, but that fact doesn't appear in my textbook. $\endgroup$ – TinaBelcher Oct 10 '14 at 1:18
  • $\begingroup$ @TinaBelcher: Yes- you are right. These facts can be found in some functional analysis textbook. $\endgroup$ – voldemort Oct 10 '14 at 1:37

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