Show that explicit formula for fixed-point does not depend on the initial point

Question

I am looking at an example application of the Banach fixed-point Theorem.

The problem is to find solutions to $Id(n) - A=y$ where $A \in M(n)$ and $Id(n)$ is the $n \times n$ identity matrix.

The problem can equivalently be written as $x=Ax+y$ such that the solution is a fixed-point of the function $f_y: \mathbb{R}^n \to \mathbb{R}^n, f_y(x)=Ax+y$.

By the Banach fixed-point Theorem it follows that there is a unique solution and $Id(n)-A$ is invertible.

Now since the theorem is constructive we can write down an explicit representation of $x$ as a limit.

For example, let $y$ be the initial point, then

$f_y(y)=Ay+y, f_{y}^2=A(Ay+y)+y=A^2y+Ay+y,...$

such that

$x=\lim \limits_{k \to \infty} f_{y}^{k}(y)=\sum \limits_{i=0}^{\infty} A^i y$ which we know converges.

The question I have is how I can show that this representation is independent of the initial point. I know that this is clear from the proof since for any initial point we get a Cauchy sequence and the fixed-point is unique, but I'd like to see it for this example.

For this I've tried to replace the initial value $y$ by $y+c$ where $c$ is an arbitrary vector. After that I repeatedly apply the function $f$ to the previous value.

This leads to

$\lim \limits_{k \to \infty} f_{y}^{k} (y+c)=\sum \limits_{i=0}^{\infty} A^i (y+c)=x+A^k c$

and I need to show that the last term is $0$.

Am I on the right track here? How can I finish this proof please?

Thanks very much!

Edit:

I forgot to state that the matrix norm $\|A\|_{op}<1$.

Using this I found a neat way to do it. Let $\|A\|_{op}=\lambda$, then

$\|A^k c\| \leq \|A\|_{op} \|A^{k-1} c\| \leq \|A\|^k \|c\|=\lambda^k \|c\|$.

Since the operator norm is non-negative, $\lim \limits_{k \to \infty} A^k c=0$ which proves the result.

Answer 1

Let $\|A\|_{op}=\lambda$, then

$\|A^k c\| \leq \|A\|_{op} \|A^{k-1} c\| \leq \|A\|^k \|c\|=\lambda^k \|c\|$.

Since the operator norm is non-negative, $\lim \limits_{k \to \infty} A^k c=0$ which proves the result.

Answer 2

I'll answer for the scalar case

$$x_{n+1}=ax_n+b,$$ and you can generalize.

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We define

$$x_\infty:=\frac b{1-a}$$ and this allows us to write

$$x_{n+1}-x_\infty=a(x_n-x_\infty)$$ because $a-ax_\infty=b$. Then by induction

$$x_n-x_\infty=a^n\left(x_0-x_\infty\right).$$

Hence if $|a|<1$, the RHS vanishes and $x_n$ converges to $x_\infty$.