```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.> > > > quasiYou get (|Fx|p)/(|x|p) = 1, but the book says:|x|p=|Fx|p implies |F|p>=1I cannot get |F|p>=1This is from a formal publish book. It does not seems a typo.Thanks
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