The following is a part of a theorem in Murphy's C*-algebras and operator theory:
Let $A$ be a C*-algebra acting irreducibly on a Hilbert space $H$ and $q$ be a nonzero projection in $A$ of minimal finite rank. Show that $q$ is rank one.
Murphy uses below argument for it:
While I think $qAq=\Bbb q$ implies that $q$ is rank one operator, because if $q=x\otimes x + y\otimes y $ for $x,y\in H$, then for $u\in A$, $$(x\otimes x+ y\otimes y) u(x\otimes x+ y\otimes y)= \lambda_1 x\otimes x + \lambda_2 x\otimes y + \lambda_3 y\otimes x + \lambda_4 y\otimes y \not \in \Bbb Cq $$ Is my argument correct? Where is my mistake?
Also in the earlier part of this book, The writer supposes $\phi:K(H) \to K(H')$ is an $*-$ isomopphism and $p_e = e\otimes e \in K(H)$ and put $q_e=\phi(p_e)$. He shows that $q_e K(H) q_e = \Bbb C q_e$. Then he claims this is easily inferred from this that $q_e$ is rank one projection. BUt I do not know how he concludes it. Please give me a hint. Thanks
It is not true that $qAq=\mathbb{C}q$ implies $\operatorname{rank}(q)=1$, if the action of $A$ is not irreducible. See the following counterexample.
Consider the $C^*-$algebra : $A=\{cId_{2\times 2}$, $c\in\mathbb{C}\}$.
Let $q=Id_{2\times 2}$. Notice that $q$ is a projection and $\operatorname{rank}(q)=2$ and $qAq=\mathbb{C}q$. Notice that the action of $A$ is not irreducible on $\mathbb{C}^2$.