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I want to show that $I-A$ is nonsingular if $\Vert A \Vert_p < 1$, where $\Vert \cdot \Vert_p : \mathbb{R}^{n \times n} \rightarrow \mathbb{R}$ is defined as $$\max_{x \in \mathbb{R}^n,\, \Vert x \Vert_p = 1} \Vert Ax \Vert_p.$$ Here's my attempt: Assume towards contradiction that $\det(I-A) = 0$. This means that there exists at least one eigenvalue $\lambda$ of $I-A$ such that $\lambda = 0$. Since the eigenvalues of the identity matrix $I$ are all $1$'s, by linearity, $A$ has at least one eigenvalue $\lambda = 1 \in \sigma(A)$. We can think of $Ax$ as a linear transformation on $x$ that stretches/shrinks $x$ in each of $A$'s eigenvector's direction by the corresponding eigenvalue. We're given that $\Vert x \Vert_p = 1$, so $\Vert Ax \Vert_p < 1$ means that $\max \sigma(A) < 1$. So we have a contradiction and can conclude that $I-A$ is nonsingular.

I think I'm pretty close, but I just feel a little iffy about the "eigenvalue stretch/shrink" argument, since for example, if we have a $2 \times 2$ rotation matrix $A$, it doesn't even have an eigenvalue -- although in this case, $\Vert A \Vert_p = 1$ so this particular example doesn't really apply... Could anyone confirm if my attempt is complete or point out anything that I'm missing? Thanks a lot!

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    $\begingroup$ Does this answer your question? Why does $(I-A)$ has inverse when $\|A \|< 1$ $\endgroup$
    – Conifold
    Sep 22 '20 at 6:14
  • $\begingroup$ Yeah I noticed that post afterwards, but it provided very little detail in the step I'm unsure of (norm < 1 $\rightarrow$ max eigenvalue < 1) so I decided to keep this question. Hope that's okay! $\endgroup$
    – user594147
    Sep 22 '20 at 14:17
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I think your argument is fine. The spectrum of $A$ is a compact subset of $\Bbb C$ which is contained in a disk of radius $\|A\|$, so if $I-A$ were singular, $1$ would be an eigenvalue of $A$, leading to $1 \leq r(A) \leq \|A\| < 1$, where $r(A)$ is the spectral radius of $A$. This is essentially what you did.

One intuitive comparison is that $$\frac{1}{1-x} = \sum_{n \geq 0} x^n$$for $|x|<1$ (just a geometric series), so you want to say that $$(I-A)^{-1} = \sum_{n \geq 0} A^n.$$To make sense of the above, one needs to show that the above series on the right side converges, and now $\|A\| <1$ kicks in, since $$\left\|\sum_{n\geq 0} A^n\right\| \leq \sum_{n \geq 0}\|A\|^n = \frac{1}{1-\|A\|}<+\infty.$$In other words, not only $I-A$ is non-singular, you get for free an expression for the inverse $(I-A)^{-1}$, and also that $\|(I-A)^{-1}\| \leq (1-\|A\|)^{-1}$. This works in any unital Banach algebra.

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    $\begingroup$ Thanks for your reply! Indeed the ultimate goal of this question is to prove that $(I-A)^{-1} = (I+A)$. I proved the rest but I just wasn't sure about this particular part. I actually didn't know the spectrum of $A$ is contained in a disk of radius $\Vert A \Vert$, so thanks for pointing that out! $\endgroup$
    – user594147
    Sep 22 '20 at 21:30
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    $\begingroup$ By the way, there's this operator version of the Cauchy-Hadamard formula: $r(A) = \limsup_{n \to +\infty} \|A^n\|^{1/n}$. $\endgroup$
    – Ivo Terek
    Sep 22 '20 at 21:39
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    $\begingroup$ Also, $(I-A)^{-1} = I+A+A^2+\cdots$, not $I+A$ (telescope). $\endgroup$
    – Ivo Terek
    Sep 22 '20 at 23:16
  • $\begingroup$ Sorry, yes that was for the first part of the question where $A$ is positive definite and $\Vert A \Vert_p < 1$, I misread. Here we're dealing with general square matrices over the reals so we ultimately want to show $(I-A)^{-1} = \sum_{k=0}^{\infty} A^k$. $\endgroup$
    – user594147
    Sep 22 '20 at 23:56
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Another approach I do like and pheraps more general is the following, using Neumann series

Theorem : If $H \in L(E)$ with $E$ Banach space and $\vert \vert H \vert \vert < 1$ then $I-H$ is invertible. Besideds $\sum\limits_{0}^{\infty}H^j$ is normally convergent to $(I-H)^{-1}$.

Proof : It holds that $\sum\limits_{0}^{\infty}\vert \vert H^{j} \vert \vert \leq \sum\limits_{0}^{\infty}\vert \vert H \vert \vert^{j}$, so we have the convergence since the RHS is a geometric series of ratio less than $1$. Let's verify that this sum is $(I-H)^{-1}$ indeed.

Let's $L = \sum\limits_{j\geq 0} H^{j}$ it holds that, by continuity of the right multiplication operator for a matrix, $LH=(\sum\limits_{0}^{\infty}H^j)H=L-I$, which algebricallyleads to the thesis.

$L(E)$ the Banach space of linear continuos application from $E \to E$

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