Why is the Calkin algebra purely infinite? I tried using the fact that in a simple unital $C^*$-algebra, $\mathcal{A}$, purely infinite is equivalent to the following: If $x\in\mathcal{A}$ is non-zero, then there exists $a,b\in\mathcal{A}$ with $axb=1$.
 A: It is well known that $\mathcal{K(H)}$ is the only non-trivial closed ideal of $\mathcal{B(H)}$ whenever $\mathcal{H}$ is separable and infinite-dimensional. By correspondence, it is clear that $\mathcal{Q(H)}$ is simple. 
Let $T\in\mathcal{B(H)}$ be non-compact. Our goal is to find elements $A,B\in\mathcal{B(H)}$ such that $\pi(I)=\pi(A)\pi(T)\pi(B)$, where $\pi:\mathcal{B(H)}\rightarrow \mathcal{Q(H)}$ is the canonical quotient map. With this in mind, we may assume that $T\geq 0$ and $\|T\|=1$ (by replacing $T$ by $\frac{T^*T}{\|T^*T\|}$ if necessary). There exists a spectral measure $E_T:\mathrm{Bor}(\sigma(T))\rightarrow\mathcal{B(H)}$ such that $E_T(\Delta)$ is a projection for all Borel subsets $\Delta$ of $\sigma(T)$, and $$T=\int_0^1x\,dE_T(x).$$ Since $T\notin\mathcal{K(H)}$, there exists $\varepsilon\in(0,1]$ such that $E_{T}((\varepsilon,1])$ is a projection of infinite rank. Define $\iota$ and $f$ on $[0,1]$ by $\iota(x)=x$ and $$f(x)=\left\{\begin{array}{lr}0,&0\leq x\leq\varepsilon\\ 1/x,&\varepsilon<x\leq 1\end{array}\right.,$$ and note that the map
$\varphi:L^\infty_{\mathrm{Bor}}(\sigma(T))\rightarrow \mathcal{B(H)}$ given by $\varphi(g)=\int_0^1g(x)\,dE_T(x)$ is a $*$-homomorphism. Having said this, we observe that by defining $S:=\varphi(f)$, we have 
\begin{align*}
TS=\varphi(\iota)\varphi(f)=\varphi(\iota f)&=\int_0^1xf(x)\,dE_T(x)
=\int_\varepsilon^1 dE_T(x)
=E_T((\varepsilon,1]).
\end{align*}
Hence, by replacing $T$ by $TS$ if necessary, we may assume that $T$ is a projection of infinite rank. In this case, $T$ is equivalent to $I$, and so there is a partial isometry $U$ such that $U^*U=T$ and $UU^*=I$. But then $I=UU^*UU^*=UTU^*$ and we see that 
$$\pi(I)=\pi(U)\pi(T)\pi(U)^*.$$ By the remarks stated in the question, we deduce that $\mathcal{Q(H)}$ is purely infinite.
A: We could have also used the following facts:
(1) A simple $C^*$-algebra is purely infinite if and only if it has real rank zero and every non-zero projection is infinite.
(2) If $\mathcal{A}$ is a $C^*$-algebra and $I$ is a closed ideal of $\mathcal{A}$, then the real rank of $\mathcal{A}/I$ is zero whenever the real rank of $\mathcal{A}$ is zero. 
Let's put the pieces together. 
We know that $\mathcal{B(H)}$ is a von Neumann algebra and hence has real rank zero (as the Borel functional calculus ensures that any self-adjoint element can be approximated by an invertible self-adjoint). Thus, (2) implies that the real rank of the Calkin algebra $\mathcal{Q(H)}=\mathcal{B(H)}/\mathcal{K(H)}$ is also zero. Of course, the Calkin algebra is simple (when $\mathcal{H}$ is separable) and has no finite projections (any finite projection in $\mathcal{B(H)}$ belongs to $\mathcal{K(H)}$ and hence disappears in the quotient) and therefore the result follows from (1).
