TITLE: Why is the following operator invertible
QUESTION [6 upvotes]: Suppose that $T : X \rightarrow X$ is an isometry between two Banach spaces. Then if $|\lambda| < 1$, then $T^{-1} - \frac{1}{\lambda}I_X$ is invertible. Can someone explain this step as it is critical step in one of my proofs.

REPLY [6 votes]: Fix $v \in X$. We want to show there is a unique $w \in X$ such that $\displaystyle T^{-1}w - \frac{1}{\lambda}w=v$ (this shows bijectivity, and by the bounded inverse theorem, the inverse is bounded as well). 
This is equivalent to finding a unique $w$ such that $\lambda T^{-1}w -\lambda v = w$. Remember, $v$ is fixed.
So define $A: X \to X$ by $Ax = \lambda T^{-1}x -\lambda v$.
Then $\|Ax-Ay\| = |\lambda|\|T^{-1}x - T^{-1}y\|\le |\lambda|\|T^{-1}\|\|x-y\|.$
Now since $T$ is an isometry so is $T^{-1}$, and so its operator norm is $1$. Thus $\|Ax-Ay\| \le |\lambda|\|x-y\|$ for $|\lambda|<1$. Hence $A$ is a contraction. Since $X$ is a Banach space, we can apply the contraction mapping theorem to get there is a fixed point $Aw = w$. This means $\lambda T^{-1} w -w = \lambda v$, or, $\displaystyle (T^{-1} - \frac{1}{\lambda}I)w =v$. Hence $T^{-1} - \frac{1}{\lambda}I$ is invertible.