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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, $|A|<1$?

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    $\begingroup$ What norm is used on the matrices? $\endgroup$
    – mvw
    Jul 15, 2014 at 9:05
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    $\begingroup$ @mvw For some stupid reason I assumed $|A|$ meant the determinant of the matrix A, but I just looked it up and the book is using the Frobenius Norm. $\endgroup$ Jul 15, 2014 at 9:13
  • $\begingroup$ Ok. I just expected the proof to be very similiar to the proof for a geometric series of numbers. $\endgroup$
    – mvw
    Jul 15, 2014 at 9:16

3 Answers 3

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Since $|A|<1$, and since you state that $\left|A^k\right|\le|A|^k$, you clearly have $\lim_{k\to\infty}\left|A^k\right|\to0$.

I don't know which norm you are using, but for every norm this also means $\lim_{k\to\infty}A^k\to0$.

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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.

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  • $\begingroup$ I am very interested in your last statement (the convergence in case the norm if more than 1) could you indicate me some resource? $\endgroup$
    – pinpon
    Feb 12, 2020 at 15:58
  • $\begingroup$ @pinpon For the result with "$\epsilon$", see lemma 11 in math.drexel.edu/~foucart/TeachingFiles/F12/M504Lect6.pdf For the last line, choose $A$ nilpotent with $||A||>1$. $\endgroup$
    – user91684
    Feb 16, 2020 at 9:32
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Note that

if a series $\sum_{k=0}^{\infty}a_n$ converges then $\lim_{n\to \infty} a_n =0$.

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