Matrix series question

Question

Let $A\in\mathbb{R}^{n\times n}$. When $\sum_{k=0}^{\infty}A^k$ converge, there exists a norm such that, for any positive integer $K$, $$\left\|(I-A)^{-1}-\sum_{k=0}^m A^k\right\| \le \dfrac{\|A\|^{m+1}}{1-\|A\|}.$$ So far my work is I know $\sum_{k=0}^{\infty}A^k=(I-A)^{-1}$, then by rearrangement, $$(I-A)^{-1}-\sum_{k=0}^{m}A^k=\sum_{k=m+1}^{\infty}A^k$$For any norm I can claim that $$\left\|(I-A)^{-1}-\sum_{k=0}^{m}A^k\right\|\le\sum_{k=m+1}^{\infty}\|A\|^k.$$ Then I want to claim something like $$\|(I-A)^{-1}\|\le \dfrac{1}{1-\|A\|},$$ but I don't think I can plug this in. Does anyone have any ideas?

Answer 1

As you have correctly stated, we have $$ \|(I-A)^{-1}-\sum_{k=0}^{m}A^k\|\le\sum_{k=m+1}^{\infty}\|A\|^k. $$ The sum on the right is a geometric series. We have $$ \sum_{k=m+1}^{\infty}\|A\|^k = \|A\|^{m+1} \cdot \sum_{k=0}^{\infty}\|A\|^k = \frac{\|A\|^{m+1}}{1 - \|A\|}. $$ So, we indeed have the desired result, namely $$ \|(I-A)^{-1}-\sum_{k=0}^{m}A^k\|\le \frac{\|A\|^{m+1}}{1 - \|A\|}, $$ which holds for any (sub-multiplicative) matrix norm $\|\cdot\|$ if $\|A\| < 1$.

An important step, then, is to show that if $\sum_k A^k$ converges, then there must exist a matrix norm for which $\|A\| < 1$. In order to do this, consider the relationship between $\|A\|$ and the eigenvalues of $A$.