Show that the operator is invertible Let $T:X \rightarrow X$ be a bounded linear operator ($X$ is a Banach space) and $ ||T|| < 1 $. Show that the operator $$I-T$$ is invertible, where $I$ denotes the identity operator.
Hint. How can you represent $\frac{1}{1-z}$ when $z$ is a complex number such that $|z|<1$?
Invertibility requires us to show that there exists an $(I-T)^{-1}$ operator such that composing it with $(I-T)$ yields $I$.
I think the hint I'm provided with suggests that $\frac{1}{1-z}$ is a sum of an infinite series $\sum_{n=0}^{\infty}z^{n}$ but how am I supposed to use this fact here?
I'd be grateful for a detailed explanation as funcional analysis is a hard nut to crack for me. 
 A: The space of bounded linear operators on $X$, denoted $\mathcal{L}(X)$, is a Banach space, since $X$ is complete. The norm associated with $\mathcal{L}(X)$ is the operator norm, i.e.
\begin{align*}
\|T\| = \sup_{\|x\| = 1} \|Tx\|.
\end{align*}
Recall that a normed vector space is complete if and only if every absolutely convergent series converges. 
So consider the series
\begin{align*}
\sum_{n=0}^{\infty} \|T\|^n
\end{align*}
Since $\|T\| < 1$, this series converges, hence
\begin{align*}
\sum_{n=0}^{\infty} T^n
\end{align*}
converges in $\mathcal{L}(X)$ to some element, call it $S$. To show that $S = (I-T)^{-1}$, it suffices to show that $S(I-T) = I$. Observe,
\begin{align*}
S(I - T) &= \sum_{n=0}^{\infty}  T^n (I - T) \\
&= \sum_{n=0}^{\infty}  T^n - \sum_{n=0}^{\infty} T^{n+1} \\
&= \sum_{n=0}^{\infty}  T^n - \sum_{n=1}^{\infty} T^{n} \\
&= T^0 \\
&= I
\end{align*}
Thus, $S = (I-T)^{-1}$.
A: When the operator $a*I_d - b ad(Y)$ is invertible, $a$, $b$ - complex numbers, $I_d$ - identity operator, $ad(Y)$ - adjoint operator generated by matrix $Y$. All acts in some linear finite dimensional vector space. By "when" I mean necessary and sufficient conditions on $a$, $b$, $Y$.
