# showing $Ax=b$ has a unique solution by finding the fixed points of a function

Let $n ∈ \mathbb N$. Consider an $(n×n)$-matrix A with real components and a column vector $b ∈ \mathbb R^n$. They give rise to an affine transformation $T : \mathbb R^n → \mathbb R^n$ with $T(x) = Ax+b$. Consider the Euclidean metric on $\mathbb R^n$:

Let I be the identity $(n × n)$-matrix. Suppose that $\|I − A\| < 1$. Show that the equation $Ax = b$ has a unique solution. Conclude that A is invertible.

My idea is to find fixed points of $T(x)=(I-A)x+b$. How can I do that?

• Have you already heard of the Banach fixed point theorem? Nov 15, 2015 at 13:19
• IF it is the same as the Contraction Mapping Principle, yes I have heard! Nov 15, 2015 at 13:21
• Yes, it's the same thing. Can you see how to apply it here? Nov 15, 2015 at 13:23
• I am not sure ! Nov 15, 2015 at 13:27
• It's done in the answer below. Is there any specific point that you don't understand in the answer? Nov 15, 2015 at 13:33

Use the Banach fixed point theorem. The map $T \colon \mathbb{R}^n \rightarrow \mathbb{R}^n$ is a contraction because
$$||T(x) - T(y)|| = ||(I - A)(x - y)|| \leq ||I - A|| \cdot ||x - y||$$
and $||I - A|| < 1$ (here, I assume you are working with the operator norm). The Banach fixed point theorem then implies the existence of a unique fixed point, showing that $Ax = b$ has a unique solution and thus $A$ is invertible.