Idempotent elements in a C^*- algebra

Question

While reading the proposition above, I'm stuck on understanding the three parts below.

1. Why is $z$ invertible?

2. How did we get $ez= ee^*e$?

3. Why is $(1-te+tp)$ invertible?

Thank you in advance!

Answer 1

1. Setting $a = (e^*-e)$, we have that $a^*a\geq 0$, so the spectrum of $a^*a$ is contained in $[0, \infty )$. By the spectral mapping Theorem, the spectrum of $1+a^*a$ is contained in $$ 1+[0, \infty ) = [1, \infty ). $$ Since $0$ is not in the latter set, it follows that $1+a^*a = 1+ (e-e^*)(e^*-e)$ is invertible.

2. This is easy. Just multiply it out and use that $e^2=e$.

3. Observe that the element $n=e-p$ satisfies $n^2=0$ (use the relations $ep=p$, and $pe=e$ to prove this). So $$ (1-tn)(1+tn) = 1-t^2n^2=1, $$ whence $1-tn = 1-te+tp$ is invertible.