How do i show an iteration using Newtons method in $\mathbb{R}^m$?

Question

My task is this:

Let $\textbf{F}:\mathbb{R}^m \to \mathbb{R}^m$. A fixpoint for $\textbf{F}$ is the same as a zero for $\textbf{G}(\textbf{x}) = \textbf{F}(\textbf{x}) - \textbf{x}$. Show that when we use Newtons method on $\textbf{G}$ we get the iteration: $$\textbf{x}_{n + 1} =\left(\textbf{F}'(\textbf{x}_n) - \textbf{I}_n)\right)^{-1}\left(\textbf{F}'(\textbf{x}_n)\textbf{x}_n - \textbf{F}(\textbf{x}_n)\right).$$

My work so far:

We observe that $\textbf{G}'(\textbf{x}) = \textbf{F}'(\textbf{x}) - \textbf{I}$ (think about why we get $\textbf{I}$). Which gives us;

$\textbf{x}_{n + 1} = \textbf{x}_n - \textbf{G}'(\textbf{x}_n)^{-1}\textbf{G}(\textbf{x}_n) = \textbf{x}_n - \left(\textbf{F}'(\textbf{x}_n) - \textbf{I}_n)\right)^{-1}\big( \textbf{F}(\textbf{x}_n) - \textbf{x}_n\big)$.

I am stuck here and not sure how to finish. I may have done some of my calculations wrong.

Any help/hints are more than welcome!

Answer 1

After you apply Newton's method to get $$ x_{n+1} = x_n - G'(x_n)^{-1}G(x_n), $$ factor out a $G'(x_n)^{-1}$ to get $$ x_{n+1} = G'(x_n)^{-1} \left( G'(x_n)x_n - G(x_n) \right) $$

Now substitute in your formulas for $G$ and $G'$.