Existence of inverses of linear combinations of bounded operators

Question

If $V$ is a banach space and $T$ is a bounded linear operator on $V$. If $\|T\|< a$ where $a>0$, then is $A=T-aI$ invertible? If so why? We know that if $\|T\|<1$ then $(I-T)$ is invertible.

Answer 1

Note $$ T-aI=-a(I-\frac{1}{a}T) $$ and by assumption $$ \left|\left| \frac{1}{a}T \right|\right|=\frac1{|a|}||T||<1 $$ So $(I-\frac1aT)^{-1}$ is invertible and the inverse of $T-aI$ is then $$ -\frac1a(I-\frac1aT)^{-1} $$