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|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