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This question already has an answer here:

Let $A \in \mathbb{R}^{n \times n}$ be a real $n \times n$ matrix such that $$\left\lVert I - A \right\rVert < 1$$ where $I$ is the identity matrix and $\left\lVert A \right\rVert = \sup_{x \in \mathbb{R}^n \\x \neq 0} \frac{\left\lVert Ax \right\rVert}{\left\lVert x \right\rVert}$. I need to show that $A$ is invertible.

I'm not sure how to proceed here. I was thinking that since $\left\lVert I - A \right\rVert < 1$, I could use the fact that $A$ is a Lipschitz continuous function with Lipschitz constant $\left\lVert A \right\rVert_{\text{Lip}} < 1$, i.e., $A$ is a contraction. Since $A$ is a contraction and $\mathbb{R}^{n \times n}$ is complete, then by the Banach Fixed Point Theorem, $A$ has a fixed point. But I'm not sure how this would show that $A$ is invertible or if this is the right track.

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marked as duplicate by user99914, Math1000, Error 404, Jonas Meyer real-analysis Oct 29 '17 at 5:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Every linear operator has a fixed point! And you only need think about the fact that 1 can't be an eigenvalue for $I-A$. $\endgroup$ – Jonas Meyer Oct 29 '17 at 5:41
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Lemma: Let $B \in \mathbb{R}^{n \times n}$. If $\Vert B \Vert <1$, then $I-B$ is invertible and $$(I-B)^{-1}=\sum_{k=0}^{\infty} B^k.$$

Proof: Since $\Vert B^k \Vert \leq \Vert B \Vert^k$, $\forall k \in \mathbb{Z}^+$, then the series $$S=\sum_{k=0}^{\infty} B^k$$ is absolutely convergent (hence, convergent). Let $S_m:=\displaystyle \sum_{k=0}^m B^k$ (observe that $(S_m)$ is a convergent sequence and $S_m \to S$). Notice that $$(I-B)S_m=\sum_{k=0}^m (B^k-B^{k+1})=I-B^{m+1}.$$ Also, $$\Vert (I-B^{m+1})-I \Vert=\Vert B^{m+1}\Vert \leq \Vert B \Vert^{m+1} \to 0$$ when $m \to \infty$. Therefore $$(I-B)S=\lim_{m \to \infty} (I-B)S_m=I.$$ The proof of $S(I-B)=I$ is analougous. $\mathtt{Q.E.D}$

Just put $B:=I-A$ and you're done!

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Set

$B = I - A; \tag 1$

then

$\Vert B \Vert = \Vert I - A \Vert < 1; \tag 2$

thus the infinite sequence

$S_n = \sum_0^n B^i \tag 3$

satisfies, for $m > n$,

$\Vert S_m - S_n \Vert = \Vert \sum_{n + 1}^m B^i \Vert \le \sum_{n + 1}^m \Vert B \Vert^i = \Vert B \Vert^{n + 1} \sum_0^{m - n- 1} \Vert B \Vert^i \le \Vert B \Vert^{n + 1} \sum_0^\infty \Vert B \Vert^i = \dfrac{\Vert B \Vert^{n + 1}}{1 - \Vert B \Vert}, \tag 4$

which shows that for $n$ sufficiently large, $\Vert S_n - S_m \Vert$ may be made arbitrarily small; thus the sequence $S_n$ is Cauchy and hence has a limit. Since

$(I - B) \sum_0^n B^i = I - B^{n + 1}, \tag 5$

letting $n \to \infty$ we conclude that

$(I - B)\sum_0^\infty B^i = I, \tag 6$

from which it follows that $A = 1 - B$ is invertible with inverse

$\sum_0^\infty B^i = \sum_0^\infty (I - A)^i. \tag 7$

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