Proof of "$\| I-A\|<1$ implies $A$ is invertible" using contraction mapping For a linear operator $A:\mathbb{R}^n\rightarrow \mathbb{R}^n$ consider its norm defined by $$\| A\| = \sup\{ |Ax| \,\,|\,\, x\in\mathbb{R}^n, \,\, \|x\|\le 1\}.$$ From the definition it is clear that $\| Ax\| \le \|A\| \|x\|$ for all $x\in\mathbb{R}^n$. With this notation, we come to the problem. 
Theorem: If $A$ is a linear operator with $\| I-A\|<1$ then $A$ is invertible. 
Proof: Let $\| I-A\| = c<1$. We can assume that $A\neq I$ (o.w. we are done), hence $c>0$. Let us denote the operator $I-A$ by $T$. Notice that if $\|x\| \le 1$ then $\| Tx\| \le \| T\| \|x\| <1$. Hence, $T$ takes closed unit ball of $\mathbb{R}^n$ into itself. Moreover, $\|Tx-Ty\|  \le \|T\| \| x-y\| \le c\| x-y\|$ where $0<c<1$. This means that $T$ is a contraction mapping on closed unit ball; it must have unique fixed point. But $0$ is already a fixed point of $T$; thus for $x\neq 0$ we have $T(x)\neq x$ i.e. $(I-A)(x)\neq x$ i.e. $A(x)\neq 0$, i.e. $A$ is injective on closed unit ball (and hence on $\mathbb{R}^n$). This forces that $A$ must be invertible. 
Q. Is this proof correct? (I tried to use property of contraction of mapping, which we got from hypothesis). 
 A: Yes, this is entirely correct. However, this proof works only in finite dimensions, since an injective linear operator on an infinite dimensional Banach space is no longer necessarily invertible.
The idea can be made to work even in infinite dimensions, though. Let $X$ be a real Banach space and $A \in \mathcal B(X)$ be such that $\lVert I - A\rVert = c < 1$. Define $\mathcal F_y (x) = y + (I-A)x$. By much the same argument as before, $\mathcal F_y$ is a contraction and has a unique fixed point $x_y$, for which $x_y = y + (I-A)x_y$ holds, or equivalently $Ax_y = y$. Setting $B y = x_y$, we have just proven that $AB = I$, and since $BAx$ is the unique fixed point of the mapping $t \rightarrow Ax + (I-A)t$ (which clearly is given by $x$), we also have $BA = I$ and therefore, $B = A^{-1}$. By the triangle inequality, $x_y = y + (I-A)x_y$ implies $\lVert x_y \rVert \leq \lVert y \rVert + \lVert (I-A) \rVert \lVert x_y \rVert \leq \lVert y \rVert + c \lVert x_y \rVert$. Thus, $\lVert x_y \rVert \leq \frac{1}{1-c} \lVert y \rVert$ and hence $\lVert B \rVert \leq \frac{1}{1-c}$.
