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, $$\|(I-A)^{-1}\|\ge\frac{1}{\|I-A\|}\ge\frac{1}{1+\|A\|}$$


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