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Show how to use an infinite series of matrix powers to compute $A^{-1}$ for $A \in \mathbb{R}^{n\times n}$. Which conditions must be satisfied by the eigenvalues of $A$ for the series to converge?

MathSchoolar
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  • A hint to get you started: for $x\in (-1,1)$ we have $\frac{1}{1-x} = \sum_{n=0}^\infty x^n$. – James Nov 07 '17 at 15:50
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    I already found [this explatation](https://math.stackexchange.com/a/992330/152622). So I need to find a $B$ so that $A=(I-B)$, $\|B\|<1$ and then the cauchy sequence converges to $A^{-1}$. I am just not sure how to formaly proof this. – MathSchoolar Nov 07 '17 at 16:04
  • @Math1000 I think you meant the reverse inequality. – Federico Poloni Nov 07 '17 at 16:32
  • Indeed, $\|A\|\geqslant |\lambda|$ for any eigenvalue $\lambda$. – Math1000 Nov 07 '17 at 16:52
  • @Math1000 where does this come from? This seems to contradict the information [here](https://math.stackexchange.com/a/242443/152622) – MathSchoolar Nov 07 '17 at 17:47
  • If $\|x\|=1$ and $Ax=\lambda x$, then $\|Ax\| = \|\lambda x\| = |\lambda|$... – Math1000 Nov 07 '17 at 18:22

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