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Banach's Lemma:

Let $C\in\mathbb R^{nxn}$ with $|C|<1$, then $I+C$ is invertible and

$\dfrac{1}{1+|C|}<|(I+C)^{-1}|<\dfrac{1}{1-|C|}$

How can we prove it? And exactly i don't understand this lemma where can it be used?

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    For a possible use of the lemma http://math.stackexchange.com/questions/1652267/proving-matrix-is-invertible-using-the-banach-lemma – 2017-01-23
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    or http://math.stackexchange.com/questions/1450042/banach-lemma-misunderstanding – 2017-01-23
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    Note that $\|C\|$ means the norm of $C$, not the absolute value of the determinant of $C$, i.e. $\|C\|$ is not $\Big|\small{|C|}\Big|$. – 2017-01-23

1 Answers 1

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Suppose $(C+I)x=0$ for some $x\in\mathbb R^n$. Then, $Cx = -x$. However, $\|C\|< 1$ implies $|Cx|<|x|$. Contradiction. Thus, $C+I$ is invertible.

Morevoer, $$\|(C+I)^{-1}(C+I)\| = 1\implies\|(C+I)^{-1}\|\ge \frac1{\|I+C\|}\ge \frac1{\|I\|+\|C\|}=\frac1{1+\|C\|}$$

Similarly, $(C+I)^{-1}C +(C+I)^{-1}= I$ implies $$\|(C+I)^{-1}\|=\|I-(C+I)^{-1}C\|\le \|I\|+\|(C+I)^{-1}\|\|C\|\implies \|(C+I)^{-1}\|\le \frac1{1-\|C\|}$$

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    not a norm. only determinant of C. Please again write accordingly :S – 2017-01-23
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    It can't be consider the diagonal $2\times2$ matrix $M$ with entries $-1, 1/2$. Clearly, the determinant of M is $-1/2$. But, $M+I$ has a column of zeroes. – 2017-01-23