Proving system of equations has unique solution using contraction mapping principle

Question

I am trying to prove that the linear system $x = Ax + b$ has a unique solution using the contraction mapping principle, where $$ A = \begin{pmatrix} \frac{1}{4} & -\frac{1}{4} & \frac{2}{15} \\ \frac{1}{4} & \frac{1}{5} & \frac{1}{2} \\ -\frac{1}{4} & \frac{1}{3} & -\frac{1}{3} \end{pmatrix} ~~\text{ and }~~ b = \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix}. $$

Toward this end, I've defined the map $f(x) = Ax + b$ with the next logical step being to prove that $\forall x,y \in R^3 \times R^3$ that $$ \|f(x) - f(y)\| = \|A(x - y)\| \le \|A\|\|x-y\| $$ and $\|A\| < 1$. But I'm not sure how to prove that $\|A\| < 1$.

One thing I notice here is that the L2-norms of the rows of $A$ are all less than one, and the L2-norms of the columns of $A$ are all less than one. Is this sufficient to prove the claim?

Answer 1

Observe here you have \begin{align} \|A(x-y)\|_2 \leq \|A\|_\text{op}\|x-y\|_2 \end{align} where \begin{align} \|A\|_\text{op} = \sup_{\|x\|_2=1}\|Ax\|_2 = \sigma_\text{max}(A) \leq \left(\sum^n_{i=1}\sum^n_{j=1}|a_{ij}|^2\right)^{1/2} =\sqrt{\frac{39}{50}}<1 \end{align} where $\sigma_\text{max}(A)$ denotes the largest singular value of $A$.

Answer 2

Note that it suffices to prove that $\|A\|<1$ for some submultiplicative norm. The norm doesn't have to be the induced $2$-norm.

Here, the sum of moduli of entries on each row of $A$ is smaller than $1$: $$ \left\{\begin{aligned} \frac14+\frac14+\frac2{15}&<\frac14+\frac14+\frac28<1,\\ \frac14+\frac15+\frac12&<\frac14+\frac14+\frac12=1,\\ \frac14+\frac13+\frac13&<\frac13+\frac13+\frac13=1. \end{aligned}\right. $$ Hence $\|A\|_\infty=\max_i\sum_j|a_{ij}|<1$ and $A$ is a contraction map.