```Date: Jan 8, 2013 5:49 PM
Author: quasi
Subject: Re: Question about linear algebra matrix p-norm

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|pClearly 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 therestriction 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 compactdomain, hence the range of g is compact. It follows thatg, and hence also f, has a maximum value. Since f(v) >= 0for all v, and since F is nonzero, the maximum value off 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.quasi
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