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fl
Posts:
89
Registered:
10/8/05


Re: Question about linear algebra matrix pnorm
Posted:
Jan 8, 2013 6:17 PM


On Tuesday, January 8, 2013 5:49:54 PM UTC5, quasi wrote: > On Tue, 08 Jan 2013 17:30:19 0500, quasi <quasi@null.set> wrote: > > > > >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). > > > > A few more details to support the above claim ... > > > > Consider the function f: (R^n)\{0} > R defined by > > > > f(v) = Fvp/vp > > > > Clearly f(v) >= 0 for all v. > > > > Let g the restriction of f to > > > > {v : v = 1} > > > > where v is the 2norm of v. In other words, g is the > > restriction of f to the standard unit (n1)sphere. > > > > Clearly, for any positive real number c, > > > > f(cv) = f(v) > > > > hence, the range of f is the same as the range of g. > > > > But g is a continuous realvalued function with a compact > > domain, hence the range of g is compact. It follows that > > g, and hence also f, has a maximum value. Since f(v) >= 0 > > for all v, and since F is nonzero, the maximum value of > > f must be positive. > > > > Therefore Fp exists and is a positive real number. > > > > >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
Thank you very much. It is clear to me now.
I do not understand the symbol \{0}. Could you explain it to me?
............. Consider the function f: (R^n)\{0} > R defined by



