# Show that $I-A$ is nonsingular if $\Vert A \Vert_p < 1$

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!

• Does this answer your question? Why does $(I-A)$ has inverse when $\|A \|< 1$ Sep 22 '20 at 6:14
• 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! Sep 22 '20 at 14:17

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.

• 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! Sep 22 '20 at 21:30
• By the way, there's this operator version of the Cauchy-Hadamard formula: $r(A) = \limsup_{n \to +\infty} \|A^n\|^{1/n}$. Sep 22 '20 at 21:39
• Also, $(I-A)^{-1} = I+A+A^2+\cdots$, not $I+A$ (telescope). Sep 22 '20 at 23:16
• 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 $(I-A)^{-1} = \sum_{k=0}^{\infty} A^k$. Sep 22 '20 at 23:56

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 $$E \to E$$