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If spectral radius $\rho(A)<1$, does the inequality $||(I-A)^{-1}||_{2} \leq 1/(1-\||A||_{2})$ hold true?

If it is correct can somebody give me link to the proof for this inequality?

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  • $\begingroup$ no, you need for example that $\|A\| < 1$ (with $\|A\| = \max_{\|v\|=1} \|Av\|$ which is also the largest singular value) for saying that $I-A$ is invertible. note that if $A$ is self-adjoint then $\|A\| = \rho(A)$ and then yes what you wrote is true. $\endgroup$
    – reuns
    Commented Apr 14, 2016 at 15:15

2 Answers 2

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No, consider $$A=\left[\begin{array}{cc}4/5& 4/5\\0&4/5 \end{array}\right].$$ The spectral radius of $A$ is $4/5$ and its norm is $2/5\sqrt{2(3+\sqrt{5})}>1$ so the quantity on RHS of the proposed inequality is negative, an impossibility.

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As others have pointed out, your statement is false ($\rho(A) \le \|A\|$). However, if $\| A\|<1$, one can show that $$ \| (I - A)^{-1}\| \le \frac{1}{1-\|A\|}. $$

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