Date: Jan 8, 2013 1:06 PM
Author: fl
Subject: Re: Question about linear algebra matrix p-norm
On Tuesday, January 8, 2013 1:36:42 AM UTC-5, quasi wrote:

> rxjwg98@gmail.com wrote:

>

>

>

> >Hi,

>

> >I am reading a book on matrix characters. It has a lemma on

>

> >matrix p-norm. I do not understand a short explaination in

>

> >its proof part.

>

> >

>

> >The Lemma is: If F is Rnxn and |F|p<1 (p-norm of F), then

>

> >I-F is non-singular....

>

> >

>

> >In its proof part, it says: Suppose I-F is singular. It

>

> >follows that (I-F)x=0 for some nonzero x. But then

>

> >|x|p=|Fx|p implies |F|p>=1, a contradiction. Thus, I-F

>

> >is nonsingular.

>

> >

>

> >My question is about how it gets:

>

> >But then |x|p=|Fx|p implies |F|p>=1

>

> >

>

> >Could you tell me that? Thanks a lot

>

>

>

> It's an immediate consequence of the definition of the matrix

>

> p-norm. By definition,

>

>

>

> <http://en.wikipedia.org/wiki/Matrix_norm>

>

>

>

> |F|p = max (|Fx|p)/(|x|p)

>

>

>

> where the maximum is taken over all nonzero vectors x.

>

>

>

> Thus, |F|p < 1 implies

>

>

>

> (|Fx|p)/(|x|p) < 1 for all nonzero vectors x,

>

>

>

> But if I - F was singular, then, as you indicate, F would have

>

> a nonzero fixed point x, say.

>

>

>

> Then

>

>

>

> Fx = x

>

> => |Fx|p = |x|p

>

> => (|Fx|p)/(|x|p) = 1,

>

>

>

> contradiction.

>

>

>

> quasi

You get

(|Fx|p)/(|x|p) = 1,

but the book says:

|x|p=|Fx|p implies |F|p>=1

I cannot get

|F|p>=1

This is from a formal publish book. It does not seems a typo.

Thanks