Show that a certain projection is rank one.

Question

Consider the following fragment from Murphy's "$C^*$-algebras and operator theory", namely a part of the proof of Theorem 2.4.8.

Can someone explain why $q_e$ is a rank-one projection? (see marked text). Thanks in advance.

Answer 1

Let $V$ be a closed subspace of a Hilbert space $H$ and let $q$ be the orthogonal projection onto $V$. Then $qK(H)q= K(V)$ where $K(V)$ is embedded into $B(H)$ by extending maps to be $0$ on the orthogonal complement of $V$, this preserves compactness of the map.

By definition $q u q = u$ for all $u\in K(V)$ once you have embedded $u$ into $B(H)$, so it follows that $qK(H)q \supseteq K(V)$. This partial result already implies what is needed: If $q_e$ is not rank $1$ (and not $0$) then $\mathrm{dim}_{\Bbb C}F(\mathrm{im}(q_e))≥2$, which is in contradiction to: $$\Bbb Cq_e=q_eK(H')q_e \supseteq K(\mathrm{im}(q_e)) \supseteq F(\mathrm{im}(q_e)).$$

For completeness the remaining direction $qK(H)q \subseteq K(V)$: If $quq\in qK(H)q$ then $quq$ is zero on the orthogonal complement of $V$ and also valued in $V$, so it is the extension by zero of a linear operator defined on $V$. The image of the closed unit ball under $quq$ is pre-compact in $H$ and also contained in the closed subspace $V$, so it is also pre-compact in $V$ and $quq$ is actually a compact operator when viewed as a map $V\to V$.