Applying the Cauchy Schwarz Inequality Let $A = (a_{ij})$ be an $n \times n$ real matrix, $I = (\delta_{ij})$ the $n \times n$ identity matrix, $b \in \mathbb{R}^n$. Suppose that $$\|A - I\|_2 = \left(\sum_{i=1}^n \sum_{j=1}^n (a_{ij} - \delta_{ij})^2\right)^{\frac{1}{2}} < 1$$ 1) Use the Cauchy-Schwarz Inequality to show that the map $T: \mathbb{R}^n \to \mathbb{R}^n$ defined by $T(x) = x - Ax + b$, satisfies $\rho(Tx, Ty) \leq \alpha \rho(x,y)$ for some $\alpha \in (0,1)$, where $\rho$ stands for the unique standard Euclidean distance in $\mathbb{R}^n$.

I know that the CSI in $\mathbb{R}^n$ is defined as $$\left(\sum_{i=1}^n x_i y_i\right)^2 \leq \left(\sum_{i=1}^n x_i^2\right) \left(\sum_{j=1}^n y_j^2\right)$$ but i'm not really sure how we can apply this to our problem. We have the map $T$ defined above with a certain metric property. What can we do with it?

Write $Tx = b - (A - I)x$. Then you can see that $Tx - Ty = (A - I)(y - x)$, which implies $\rho(Tx,Ty) \le \alpha \rho(x,y)$, with $\alpha = \|A - I\|_2 < 1$.
