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Topic: Question about linear algebra matrix p-norm
Replies: 6   Last Post: Jan 9, 2013 2:42 AM

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fl

Posts: 83
Registered: 10/8/05
Re: Question about linear algebra matrix p-norm
Posted: Jan 8, 2013 1:06 PM
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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



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