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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: 10,197
Registered: 7/15/05
Re: Question about linear algebra matrix p-norm
Posted: Jan 8, 2013 5:30 PM
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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).

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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