Under which conditions does the sequence $Mx^{k+1}=Nx^{k}+c$ converge to the solution of $Ax=b$?

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Which conditions for the matrix $M$ and $N$ are necessary to make the sequence $Mx^{k+1}=Nx^{k}+c$ converge to the solution of $Ax=b$? (k is the index of the sequence not a power)

And how does the conditions change when $c$ is changed by $b$? i.e : $Mx^{k+1}=Nx^{k}+b$

For the first questions, I have heard that one of the condition is that M needs to be invertible and $\rho (M^{-1}N)<1$. But I would like to find a reference for this result.

Context This question is important because it explain the famous gradient descent method to solve Linear Leas Square problems. The minimization of LLS problem is performed using gradient descent and the method converges to the exact solution "b" when some specific relations between M,N, and A are stablished.

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The case when $M$ isn't invertible is somewhat delicate. In the extreme case $M$ and $N$ are both the zero matrix and $c = 0$, then any sequence $(x^k)$ you want solves the recurrence (and some will converge to $A^{-1}b$ and others not). Let's just restrict ourselves to the case when $M$ is invertible.

If this sequence converges to $x = A^{-1}b$, then taking the limit as $k\to\infty$ of both sides gives

$$ MA^{-1}b=NA^{-1}b+c \implies c=(M-N)A^{-1}b. $$

This is a necessary condition for the sequence to converge. (You can't just set $c = b$ unless $M-N = A$.) Let us assume it holds and go forward.

Now, make a change of variables $e^k = A^{-1}b-x^k$. The sequence $(x^k)$ converges if, and only if, $(e^k)$ converges. This is the "error" of the method at the $k$th step plugging in, we see that

\begin{align*} Me^{k+1}&=MA^{-1}b-Mx^{k+1} = MA^{-1}b-(Nx^k +c) \\ &= MA^{-1}b - N(A^{-1}b-e^k)-(M-N)A^{-1}b \\ &= N e^{k}. \end{align*}

Thus $e^{k+1} = M^{-1}Ne^k$ so $e^k = (M^{-1}N)^k e^0$. For this sequence to converge for all initial conditions $e^0$, the powers of the matrix $M^{-1}N$ must converge tot he zero matrix. One can check that this is true if, and only if, $\rho(M^{-1}N) < 1$ using, for example, the Jordan canonical form.

Thus (assuming $M$ and $A$ are invertible), necessary and sufficient conditions for $(x^k)$ converging to $A^{-1}b$ for all initial conditions $x^0$ is that $c=(M-N)A^{-1}b$ and $\rho(M^{-1}N) < 1$.