# Proving that $\frac{1}{1 + \lVert\mathbf{A}\rVert} \leq \lVert (\mathbf{I}_{d} - \mathbf{A})^{-1} \rVert$

Let $$\mathbf{A} \in \mathbb{R}^{d \times d}$$ such that $$\lVert \mathbf{A}\rVert < 1$$, where $$\lVert \heartsuit\rVert$$ is matrix norm. Show that $$\begin{equation*} \frac{1}{1 + \lVert\mathbf{A}\rVert} \leq \lVert (\mathbf{I}_{d} - \mathbf{A})^{-1} \rVert \leq \frac{1}{1-\lVert\mathbf{A}\rVert}. \end{equation*}$$

I have shown that $$(\mathbf{I}_{d} - \mathbf{A})$$ is invertible, using the fact that the subset of invertiable matrices is open. Then using geometric series, I could show one side of the inequality $$\lVert (\mathbf{I}_{d} - \mathbf{A})^{-1} \rVert = \lVert\sum_{k=0}^{\infty} \mathbf{A}^{k}\rVert \leq \sum_{k=0}^{\infty} \lVert\mathbf{A}\rVert^k = \frac{1}{1-\lVert\mathbf{A}\rVert}$$

Since $$\lVert\mathbf{A}\rVert^{n+1} \leq \lVert\mathbf{A}\rVert^{n}$$ und $$\lim_{n \to \infty} \lVert\mathbf{A}\rVert^n = 0$$, we have $$\sum_{k=0}^{\infty} (-1)^k \lVert\mathbf{A}\rVert^k = \frac{1}{1+\lVert\mathbf{A}\rVert}$$ but I can't show that it bounds $$\lVert (\mathbf{I}_{d} - \mathbf{A})^{-1} \rVert$$ from below. Any help?

In general $$\|B^{-1}\|\ge\frac{1}{\|B\|}$$ since $$1=\|BB^{-1}\|\le\|B\|\|B^{-1}\|$$.
Also, $$\|I-A\|\le1+\|A\|$$. Hence combining the two, $$\|(I-A)^{-1}\|\ge\frac{1}{\|I-A\|}\ge\frac{1}{1+\|A\|}$$