Let $A=(a_{ij}) \in \mathbb R^{n \times n}$ a matrix such that $|a_{ij}|<\frac{1}{n}$ for every $i,j$. Prove that $I-A$ is invertible.
My attempt at a solution:
$I-A$ is invertible $\iff$ $(I-A)v=0$ has the trivial solution $v=0$. But $(I-A)v=0 \iff Iv-Av=0 \iff Av=v$. Let $T:\mathbb R^n \to \mathbb R^n$ be the linear transformation with $|T|_e=A$, as T is a linear transformation, $v=0$ satisfies $T(v)=v$. In order for $I-A$ to be invertible, $T$ must have a unique fixed point (as $v=0$ is a fixed point, we've already proved existence). The space $\mathbb R^{n \times n}$ is complete, so if we prove $T$ is a contraction, then, by the fixed point theorem, we can assure there is a unique point $v$ such that $T(v)=v$, and as $T$ is linear, this point has to be $0$.
I don't know how to prove $T$ is a contraction. I must prove that for $x,y \in \mathbb R^n$, $\|T(x)-T(y)\|<\alpha \|x-y\|$, with $0<\alpha <1$. I have to use the fact that $|a_{ij}|<\frac{1}{n}$, can anyone guide me on how to use this information properly?
Since $T$ is linear, it is sufficient to prove $\|Tx\| \le \alpha \|x\|$ for some $0 \le \alpha < 1$.
Also, it is sufficient to prove it is a contraction for any norm on $\mathbb R^n$, and the problem is much easier if you use the taxicab norm $\|x\| = \sum \limits_{i=1}^n |x_i|$ or the max norm $\|x\| = \max \limits_{1 \le i \le n} |x_i|$.
But if you want to use the Euclidean norm, start with $$ \|Tx\|^2 = \sum_{i=1}^n \left|\sum_{j=1}^n a_{ij}x_j\right|^2 \le n\left(\max_{1 \le i,j \le n} |a_{ij}|^2\right) \left(\sum_{j=1}^n |x_j|\right)^2 $$ Then use Cauchy-Schwarz to show $$ \sum_{j=1}^n |x_j| \le \sqrt n \|x\| .$$ You will get the result with $\alpha = n\left(\max \limits_{1 \le i,j \le n} |a_{ij}|\right)$.