Let $A$ be a C*-algebra with a nonzero minimal projection $e$.
a - Show that if $\{\pi, H\}$ is an irreducible representation of $A$ such that $\pi(e) \neq 0$, then $\pi(e)$ is a projection of rank 1.
b- Show that if $A$ is simple, then $A$ is isometrically isomorphic to $K(H)$(compact operator space) for some Hilbert space $H$.
Suppose $e$ is a nonzero minimal finite rank projection, I can prove part (a) (using the fact $\pi(eAe) = \Bbb C\pi(e)$) but in general I do not have any idea. Please give me hints for both parts. Thanks in advance.
Let $p $ be a projection with $p\leq\pi (e) $. Suppose $\pi (x_j)\to p $ weakly. As $\pi (e)p=p $, we get that $\pi (ex_je)=\pi (e)\pi (x_j)\pi (e)\to p $. But $e $ is minimal, so $ex_je=\lambda_je $ for numbers $\lambda_j $. Then $p=\lim \lambda_j\pi ( e)\in\mathbb C\,\pi (e )$, and so either $p=0$ or $p=\pi (e )$. Thus, $\pi (e) $ is rank-one.
If $A $ is simple, every representation is isometric. Let $\pi:A\to B (H) $ be irreducible. Then $\pi (A) $ is a weakly-dense simple subalgebra of $B (H) $ that contains a rank-one projection $\pi (e) $. So $\pi(A)\cap K(H)$ is a nonzero ideal of $\pi(A)$; as $\pi(A)$ is simple, $\pi(A)=\pi(A)\cap K(H)$, i.e. $\pi(A)\subset K(H)$.
It remains to show that $A=K(H)$. Let $\xi\in H$ be a unit vector in the range of $\pi(e)$. Then $\pi(e)=\langle\cdot,\xi\rangle\xi$. Now let $\eta,\nu\in H$. Since $\pi(A)''=B(H)$, by Kadison's Transitivity Theorem there exists $a,b\in A$ with $\pi(a)\xi=\eta$, $\pi(b)\xi=\nu$. Now, for $\zeta\in H$, $$ \pi(a)\pi(e)\pi(b)^*\zeta=\langle\pi(b)^*\zeta,\xi\rangle\pi(a)\xi=\langle\zeta,\pi(b)\xi\rangle\pi(a)\xi=\langle\zeta,\eta\rangle\nu. $$ This shows that the rank-one operator $\langle\cdot,\eta\rangle\nu$ is $\pi(a)\pi(e)\pi(b)^*\in \pi(A)$. So $\pi(A)$ contains all rank-one operators, and thus it contains $K(H)$. So $\pi(A)=K(H)$.