Linear algebra : Why $C_{n+1} - A^{-1} = (C_n - A^{-1})NM^{-1}$ and why $(C_n)$ converges to $A^{-1}$? Let $ A \in M_d(\mathbb{R})$ an invertible matrix. We suppose $A=M-N$ with $M \in M_d(\mathbb{R})$ is invertible. We want to approach $A^{-1}$ with a sequence of matrix $(C_n)_{n \in \mathbb{N}}$ by the recurrence relation :
$C_{n+1} = C_n + (I_d -C_n A)M^{-1}$.
We want a necessary condition to prove that $(C_n)_{n \in \mathbb{N}}$ converges to $A^{-1}$ for all choice of $C_0$.
We have the following equality $C_{n+1} - A^{-1} = (C_n - A^{-1})NM^{-1}$ that I don't understand, and we deduce $C_n - A^{-1} = (C_0 - A^{-1})(NM^{-1})^n$.
Someone could explain the two previous equals ? I begin in linear algebra so it's not intuitive to me... Thank you in advance !
 A: Observe that since
$$
C_{n+1}=C_n+(I_d-C_nA)M^{-1},
$$
we can subtract $A^{-1}$ from both sides to get
$$
C_{n+1}-A^{-1}=C_n+(I_d-C_nA)M^{-1}-A^{-1}.
$$
Since $A^{-1}A=I_d$, we can rewrite this as
$$
C_{n+1}-A^{-1}=C_n+(I_d-C_nA)A^{-1}AM^{-1}-A^{-1}.
$$
Distributing the $A^{-1}$ and regrouping results in
$$
C_{n+1}-A^{-1}=(C_n-A^{-1})+(A^{-1}-C_n)AM^{-1}.
$$
Since multiplying by $I_d$ doesn't change anything, this is the same as
$$
C_{n+1}-A^{-1}=(C_n-A^{-1})I_d+(A^{-1}-C_n)AM^{-1}.
$$
Factoring the $C_n-A^{-1}$ on the RHS gives
$$
C_{n+1}-A^{-1}=(C_n-A^{-1})(I_d-AM^{-1}).
$$
Since $A=M-N$,
$$
C_{n+1}-A^{-1}=(C_n-A^{-1})(I_d-(M-N)M^{-1}),
$$
and by multiplying this out,
$$
C_{n+1}-A^{-1}=(C_n-A^{-1})(I_d-I_d+NM^{-1})=(C_n-A^{-1})NM^{-1},
$$
as desired.
The second equality is treating this as you might deal with a geometric series/induction.  In other words, by substituting the previous line into the next line,
\begin{align}
C_0-A^{-1}&=(C_0-A^{-1})I_d=(C_0-A^{-1})(NM^{-1})^0\\
C_1-A^{-1}&=(C_0-A^{-1})NM^{-1}=(C_0-A^{-1})(NM^{-1})^1\\
C_2-A^{-1}&=(C_1-A^{-1})NM^{-1}=(C_0-A^{-1})(NM^{-1})^2\\
C_3-A^{-1}&=(C_2-A^{-1})NM^{-1}=(C_0-A^{-1})(NM^{-1})^3
\end{align}
and so on.
