quasi
Posts:
12,057
Registered:
7/15/05


Re: Question about linear algebra matrix pnorm
Posted:
Jan 8, 2013 5:49 PM


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

