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Banach's Lemma:

Let $C\in\mathbb R^{n\times n}$ 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?

Elias Costa
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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 – Emre Jan 23 '17 at 14:58
  • or http://math.stackexchange.com/questions/1450042/banach-lemma-misunderstanding – Emre Jan 23 '17 at 14:59
  • Note that $|C|$ means the norm of $C$, not the absolute value of the determinant of $C$, i.e. $|C|$ is not $\Big|\small{|C|}\Big|$. – user1551 Jan 23 '17 at 15:07
  • "Banach's Lemma" would be where you have the operator norm in there, not the determinant. – GEdgar Aug 20 '21 at 12:21

1 Answers1

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

Emre
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  • not a norm. only determinant of C. Please again write accordingly :S –  Jan 23 '17 at 16:11
  • 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. – Emre Jan 23 '17 at 16:16
  • This does not answer the question "Where is it used" – Cedric Martens Oct 26 '23 at 16:27