Find a bound for a matrix Let $A, B \in \mathbb{R}^{n\times n}$ and $I$ be an identity matrix of order $n$. Suppose $$B_k = B_{k-1} + B_{k-1} (I - A B_{k-1}), \quad (k=1,2,\ldots)$$
If $\Vert I - AB_0 \Vert = c < 1$, then show $\lim \limits_{k\to\infty} B_k = A^{-1}$ and
$$\Vert A^{-1} - B_k \Vert \leqslant \frac{c^{2^k}}{1-c} \Vert B_0 \Vert.$$
$c$ is a $\rm{constant}$.
 A: A minor remark: from the statement of the problem, one deduces that $A$ is invertible. Since $\| I-AB_0 \| < 1$, $AB_0$ is also invertible, whence it follows that $B_0$ is invertible too (but this will not be used).
Note that
$$B_k = 2 B_{k-1} - B_{k-1} A B_{k-1} ,$$
therefore
$$I - A B_k = I - 2 A B_{k-1} + A B_{k-1} A B_{k-1} = (I - A B_{k-1})^2$$
and then
$$\|I - A B_k\| \le \|I - A B_{k-1}\| ^2 \le \|I - A B_{k-2}\| ^{2^2} \le \dots \le \|I - A B_0\|^{2^k} = c^{2^k} .$$
Now,
$$\begin{equation*} \|A ^{-1} - B_k\| \le \|A^{-1}\| \cdot \|I - A B_k\| \le \|A^{-1}\| c^{2^k} . \tag{*}\end{equation*}$$
Let us prove that $\|A^{-1}\| \le \frac 1 {1-c} {\|B_0\|}$. We have
$$1 = \|I\| = \|I - A B_0 + A B_0\| \le \|I - A B_0 \| + \|A B_0\| = c + \|A B_0\| \le c + \|A\| \cdot \|B_0\| = c + \|(A^{-1}) ^{-1}\| \cdot \|B_0\| \le c + \|(A^{-1})\| ^{-1} \cdot \|B_0\|$$
therefore, rearranging terms, $\|A^{-1}\| \le \frac 1 {1-c} {\|B_0\|}$. Plugging this into $(*)$ you get $\|A ^{-1} - B_k\| \le \frac {c^{2^k}} {1-c} {\|B_0\|}$.
Finally, letting $k \to \infty$ in the above inequality, the right-hand side tends to $0$ (because $c<1$), which implies that $\lim \limits _{k \to \infty} B_k= A^{-1}$.
(In the proof we have used $\|X+Y\| \le \|X\| + \|Y\|, \space \|XY\| \le \|X\| \cdot \|Y\|$ and $\|A^{-1}\| \le \|A\|^{-1}$.)
