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I've been trying to solve this for 3 hours..

If $A$ is an $n \times n$ matrix with $\|A\|<1$ in any norm, then show that $\|(I-A)^{-1}\| \leq \frac {1}{1-\|A\|}$.

My trying is:

$$ \|(I-A)^{-1}\| = \|A^{-1} - I^{-1}\|=\| A^{-1} - I\| \leq \|A^{-1}\|-\|I\|=\|A^{-1}\||-1 \leq \|A\|^{-1} -1 $$

Multiplying both sides of the inequality by $(1-\|A\|)$, we get

$(1-\|A\|)\cdot\|(I-A)^{-1}\| \leq (1-\|A\|)\cdot(\|A||^{-1} -1)\|A\|^{-1}-1 -\|A\|^{-1}\|A\| -\|A\|= \|A\|^{-1}-\|A\|-2 \leq \|A\|^{-1}$

Therefore, $(1-\|A\|)\cdot\|(I-A)^{-1}\| \leq \|A\|^{-1}$ but $\|A\|^{-1}$ is not smaller than 1 :(

Michael Hardy
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AYARcom
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3 Answers3

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No need for infinite series: $$ (I-A)^{-1}=(I-A)^{-1}(I-A+A)=I+(I-A)^{-1}A, $$ so $$ \|(I-A)^{-1}\|\leq\|I\|+\|(I-A)^{-1}\|\|A\|=1+\|(I-A)^{-1}\|\|A\|, $$ and hence $$ (1-\|A\|)\|(I-A)^{-1}\|\leq 1. $$ Now if $\|A\|<1$, the factor on the left hand side is positive. Divide by it.

Algebraic Pavel
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If $\|\cdot\|$ is submultiplicative then the result is straightforward to show.

In particular, this gives $\|A^k\| \le \|A\|^k$.

Since $\|A\|<1$ the series $B= \sum_{k=0}^\infty A^k$ is absolutely convergent and we can easily check that $(I-A)B = 0$ and so $B = (I-A)^{-1}$.

It follows that $\|(I-A)^{-1} \| \le \sum_{k=0}^\infty \|A_k\| \le \sum_{k=0}^\infty \|A\|^k = {1 \over 1- \|A\|}$.

Batominovski
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copper.hat
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Since $|A|<1$, $(I-A)^{-1} = \sum_{k\geq 0}A^k$. By 'submultiplicativity' of the matrix norm, $$|(I-A)^{-1}|\leq\sum_{k\geq 0}|A^k|\leq\sum_{k\geq 0}|A|^k = \frac{1}{1-|A|}.$$

Francis Begbie
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