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$A$ is an $n\times n$ matrix with $\|A\|≤a<1 $ . I need to prove that the matrix $(I-A)$ is invertible with $\|(I-A)^{-1}\|\le\frac1{(1-a)}$.

It doesn't say anything more. The norm makes me confuse. How can we start to solve this. Could you please help?

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Do you know of the geometric series? –  Mariano Suárez-Alvarez Feb 17 '12 at 6:39
    
yes I know geometric series –  rose Feb 17 '12 at 6:50
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3 Answers 3

up vote 6 down vote accepted

Hint: Show that a certain series converges in the norm $\|\cdot \|$ and that this is an inverse for $I-A$.

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$\bf Hint$: Consider the series $\sum_{n=0}^\infty A^n$

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You can also argue without using the geometric series. The matrix $I-A$ is invertible if and only if $\lambda = 1$ is not an eigenvalue of $A$.

For a contradiction, assume $\lambda = 1$ is an eigenvalue. Then $Ax = x$ for some $x$ with $\|x\| = 1$, so $\|A\| \ge 1$.

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