invertible element in Calkin algebra

Question

How to conclude that $\pi(R_t)$ is invertible in $Q(H)$ in the Proposition 9.4.2?

I only know the fact that $U$ is invertible in $Q(H)$,$|A|$ is invertible in $Q(H)$.But $GL(Q(H))$ is not a convex set ,how to prove that $t|A|+(1-t)I$ is invertible in $Q(H)$?

Answer 1

As $A$ is invertible, so is $|A|$, so its spectrum is contained in $[r,\infty)$ for some $r>0$. Therefore the straight line $t|A|+(1-t)I$ from $|A|$ to the identity consists of invertible operators: the spectrum $$\sigma(t|A|+(1-t)I) = t\sigma(A) + 1-t$$ is clearly contained in $[1/2,\infty)$ for $0\leqslant t\leqslant 1/2$ and in $[r/2,\infty)$ for $1/2\leqslant t\leqslant 1$ – so it stays bounded away from zero.