Proof verification for matrix norm

Question

I am studying matrix norms from here. We are defining an operator norm on a matrix as $$\|A\|=\sup_{\substack{x\in\mathbb{C}^n}\\\|x\|=1} \|Ax\|,$$ where $\mathbb{C}$ is the set of complex numbers and vector norm is considered for $\|x\|=1$. In Proposition 4.8, it is stated that if $B$ is a matrix such that $\|B\|<1$, then $I+B$ is invertible ($I$ is an identity matrix) and $\|(I+B)^{-1}\|\le \frac{1}{1-\|B\|}$.
I tried out the proof myself as it is not stated. We can write:
$$(I+B)^{-1}=\sum_{t=0}^{\infty}(-1)^tB^t=I+\sum_{t=1}^{\infty}(-1)^tB^t=I-B[I-B+B^2-B^3+...]=I-B(I+B)^{-1}$$ $$\Rightarrow \|(I+B)^{-1}\|=\|I-B(I+B)^{-1}\|\le \|I\|+\|B\|\|(I+B)^{-1}\|=1+\|B\|\|(I+B)^{-1}\|$$ $$\Rightarrow \|(I+B)^{-1}\|\le \frac{1}{1-\|B\|}.$$ The inequality in the second line follows from sub-multiplicative property of operator norm. Can someone please let me know if this proof is fine or if there is a better proof, where can I find one. Thank you.

Answer 1

All the ideas you need are in your proof, they're just not structured in a convincing way.

I suggest that you proceed as follows: argue that because $\|B\| < 1$, the series $\sum_{t \geq 0} (-1)^tB^{t}$ converges. Then, argue that the matrix $M = \sum_{t \geq 0} (-1)^tB^{t}$ satisfies $(I + B)M = M(I + B) = I$, which allows us to conclude that $M = (I + B)^{-1}$.