If $\sen{A}<1$, then $I-A$ is invertible, and $$\bex (I-A)^{-1}=I+A+A^2+\cdots, \eex$$ aa convergent power series. This is called the Neumann series.
Solution. Since $\sen{A}<1$, $$\bex \sum_{n=0}^\infty \sen{A}^n=\frac{1}{1-\sen{A}}<\infty. \eex$$ Due to the completeness of the matrix space, $\dps{\sum_{n=0}^\infty A_n}$ converges. Since $$\bex (I-A)(I+\cdots+A^{n-1})=I-A^n, \eex$$ we may take limit to get $$\bex (I-A)\cdot \sum_{n=0}^\infty A^n=I. \eex$$
[Bhatia.Matrix Analysis.Solutions to Exercises and Problems]ExI.2.6
原文：http://www.cnblogs.com/zhangzujin/p/4105362.html