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.

>

>

>

> 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