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fl

Posts: 75
Registered: 10/8/05

Re: Question about linear algebra matrix p-norm
Posted: Jan 8, 2013 6:17 PM

Plain Text

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
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> >>>
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> >>> Fx = x
>
> >>>
>
> >>> => |Fx|p = |x|p
>
> >>>
>
> >>> => (|Fx|p)/(|x|p) = 1,
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> >>>
>
> >>> 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