```Date: Jan 8, 2013 6:17 PM
Author: fl
Subject: Re: Question about linear algebra matrix p-norm

On Tuesday, January 8, 2013 5:49:54 PM UTC-5, 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 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.> > >>> > >>You get > > >>(|Fx|p)/(|x|p) = 1, > > >>> > >>but the book says:> > >>|x|p=|Fx|p implies |F|p>=1> > >>> > >>I cannot get > > >>|F|p>=1> > >> > >Note that for any nonzero n x n matrix F over the reals, and > > >any real p >= 1, the p-norm of F exists and is a positive real > > >number (just consider the values of F restricted to the> > >standard unit (n-1)-sphere).> > > > A few more details to support the above claim ...> > > > Consider the function f: (R^n)\{0} -> R defined by > > > >    f(v) = |Fv|p/|v|p> > > > Clearly f(v) >= 0 for all v.> > > > Let g the restriction of f to> > > >    {v : |v| = 1}> > > > where |v| is the 2-norm of v. In other words, g is the> > restriction of f to the standard unit (n-1)-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 real-valued 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 |F|p exists and is a positive real number.> > > > >As observed in my previous reply, singularity of I - F implies> > >> > >   (|Fx|p)/(|x|p) = 1> > >> > >for some nonzero vector x.> > >> > >Hence the maximum value of > > >> > >   (|Fv|p)/(|v|p)> > >> > >over all nonzero vectors v must be at least one.> > >> > >Therefore |F|p >= 1.> > > > quasiThank 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
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