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



fl
Posts:
121
Registered:
10/8/05


Re: Question about linear algebra matrix pnorm
Posted:
Jan 8, 2013 1:06 PM


On Tuesday, January 8, 2013 1:36:42 AM UTC5, quasi wrote: > rxjwg98@gmail.com wrote: > > > > >Hi, > > >I am reading a book on matrix characters. It has a lemma on > > >matrix pnorm. I do not understand a short explaination in > > >its proof part. > > > > > >The Lemma is: If F is Rnxn and Fp<1 (pnorm of F), then > > >IF is nonsingular.... > > > > > >In its proof part, it says: Suppose IF is singular. It > > >follows that (IF)x=0 for some nonzero x. But then > > >xp=Fxp implies Fp>=1, a contradiction. Thus, IF > > >is nonsingular. > > > > > >My question is about how it gets: > > >But then xp=Fxp implies Fp>=1 > > > > > >Could you tell me that? Thanks a lot > > > > It's an immediate consequence of the definition of the matrix > > pnorm. By definition, > > > > <http://en.wikipedia.org/wiki/Matrix_norm> > > > > Fp = max (Fxp)/(xp) > > > > where the maximum is taken over all nonzero vectors x. > > > > Thus, Fp < 1 implies > > > > (Fxp)/(xp) < 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 > > => Fxp = xp > > => (Fxp)/(xp) = 1, > > > > contradiction. > > > > quasi
You get (Fxp)/(xp) = 1,
but the book says: xp=Fxp implies Fp>=1
I cannot get Fp>=1
This is from a formal publish book. It does not seems a typo.
Thanks



