TITLE: Geometric series of matrices
QUESTION [13 upvotes]: I am currently reading 'Vector Calculus, Linear Algebra, and Differential Forms: A Unified Approach' by J. Hubbard and B. Hubbard.  In the first chapter, there is the proposition:

Let A be a square matrix. If $|A|<1$, the series $$S=I+A+A^2+\cdots$$
  converges to $(I-A)^{-1}$.

The proof first shows that
$$S_k(I-A)=I-A^{k+1}$$
and similarly
$$(I-A)S_k=I-A^{k+1}$$
where $S_k$ is the sum of the first $k$ terms in the series.  Then it shows that
$$|A^{k+1}|\leq|A|^{k+1}$$
and according to the proof in the book, it can be said from this that $\lim_{k\to \infty}A^{k+1}=0$ when $|A|<1$.  Consequently, $S(I-A)=I$ and $(I-A)S=I$.  Therefore $S=(I-A)^{-1}$.
However I do not understand how it can be said that $\lim_{k\to \infty}A^{k+1}=0$ when $|A|<1$.  In addition, all similar propositions I have found on the internet state $|\lambda_i|<1$ as the necessary and sufficient condition.  How does this relate to the condition, $|A|<1$?

REPLY [3 votes]: It is an answer to the last question:  how does the condition (P): for every $i$, $|\lambda_i|<1$  relate to the condition, $||A||<1$ ? 
Firstly $||A||<1$ implies (P) and, in general, the converse is false.
More precisely, let $\rho(A)=\sup_i(|\lambda_i|)$ be the spectral radius of $A$. It is not true that $\rho()$ is a norm ; yet it is almost true. Indeed, for every $\epsilon$, there is a matricial norm s.t. $||A||<\rho(A)+\epsilon$ (note that the proof of the previous result is not obvious). Thus, if $\rho(A)<1$, then there is a matricial norm s.t. $||A||<1$.
Note that $\sum_{k=0}^\infty A^k$ may converge even if $||A||>1$ for a given norm.