Given a fixed point iteration $x_{n+1}=f(x_n, b)$ for solving the linear system $Ax=b,\ A\in\mathbb{R}^{n\times n}$ we define the fixed point iteration to be consistent if the solution $\hat x=A^{-1}b$ is a fixed point of $f$, meaning that $\hat x =f(\hat x, b)$. Furthermore, the fixed point iteration is linear if it is of the form $f(x,b)=Mx + Nb$.
I want to prove the following
Lemma: A linear fixed point iteration is consistent if and only if $M=\mathbb{I}- NA$ where $\mathbb{I}$ is the identity matrix.
Proof:
Given that $M=\mathbb{I}-NA$ we have $f(\hat x, b)=(\mathbb{I}-NA)\hat x + Nb=(\mathbb{I}-NA)A^{-1}b + Nb = \hat x$
Now I am stuck at the other direction: Suppose $f(\hat x, b)=\hat x$ then we have
$f(\hat x , b)=M(A^{-1}b) + Nb = A^{-1}b\\ \Rightarrow (MA^{-1})b = (A^{-1}-N)b = (A^{-1}-N)(AA^{-1})b=(\mathbb{I}-NA)\hat x = M\hat x$
But since no further assumptions about $M$ or $N$ are made I cannot conclude that $M=\mathbb{I}-NA$.
The result follows implicitly from the fact that we want the method to be consistent whatever the choice of $b$ -- or equivalently, whatever the choice of $\hat x = A^{-1}b$. Consider $M$ such that the corresponding linear fixed point iteration $f(x, b) = M x + Nb$ is consistent. For such a linear fixed point iteration, we have in particular $$ f(\hat x, b) = M \hat x + N b = (M + N A) \hat x \, . $$ Thus, if the fixed point iteration is consistent, then $$ M \hat x = (\Bbb I - N A) \hat x $$ for all $\hat x$. We have proven that $M = \Bbb I - N A$.