quasi
Posts:
12,046
Registered:
7/15/05


Re: Question about linear algebra matrix pnorm
Posted:
Jan 8, 2013 5:30 PM


fll <rxjwg98@gmail.com> wrote: >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. > >You get >(Fxp)/(xp) = 1, > >but the book says: >xp=Fxp implies Fp>=1 > >I cannot get >Fp>=1
Note that for any nonzero n x n matrix F over the reals, and any real p >= 1, the pnorm of F exists and is a positive real number (just consider the values of F restricted to the standard unit (n1)sphere).
As observed in my previous reply, singularity of I  F implies
(Fxp)/(xp) = 1
for some nonzero vector x.
Hence the maximum value of
(Fvp)/(vp)
over all nonzero vectors v must be at least one.
Therefore Fp >= 1.
quasi

