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Topic: Question about linear algebra matrix p-norm
Replies: 6   Last Post: Jan 9, 2013 2:42 AM

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quasi

Posts: 10,232
Registered: 7/15/05
Re: Question about linear algebra matrix p-norm
Posted: Jan 8, 2013 5:49 PM
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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



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