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Title: | Mathematics at DEC |
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Moderator: | RUSURE::EDP |
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Created: | Mon Feb 03 1986 |
Last Modified: | Fri Jun 06 1997 |
Last Successful Update: | Fri Jun 06 1997 |
Number of topics: | 2083 |
Total number of notes: | 14613 |
1500.0. "Normed Vector Spaces talk" by STAR::ABBASI () Tue Oct 15 1991 06:35
give a valid norm for a vector space V over real number field, where
the elements of V are m,n matrices.
a valid norm is a function (N) that maps the element in V into R (real
field) such that , for x,y in V , and 'a' is a scalar in R :
1. N(x) >= 0
2. N(x) = 0 iff x=0
3. N(ax) = |a| N(x)
4. N(x+y) <= N(x) + N(y)
one example i see now is N(x) to be the absolute value of determinant. but
i have to assume that n=m which is not general.
so what i did, is say that n,m matrix is like q sized vector where
q= m*n . (ie. traverse matrix left to right, top to bottom) and the norm
in this case is
- m*n - 1/p
| ----- p |
N(x) = | \ | x | |
p | / k |
| ----- |
- k=1 -
this, i can show easly that it meets the above 4 conditions for valid
norm for any p value 1..oo
note that for p=2 we get the "normal" length of vector. for p=oo
things looks funny, like a circle becomes a square etc.. for p=0
undefined. and p not negative.
but im sure that there is some function that can be applied
to an m,n matrix that return an element in R s.t. the above conditions
are met without me "looking" at the matrix as a vector.
i cant see it know. any one know such a function?
i've allready submitted the homework, will find out next week what is
the answer, and will post teacher's answer if different than anyones,
meanwhile you can have fun with this.
/nasser
T.R | Title | User | Personal Name | Date | Lines |
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1500.1 | this look ok. there should be more | STAR::ABBASI | | Tue Oct 15 1991 06:43 | 9 |
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i think i just saw one function
N(x) = total sum of absolute value of elements of matrix.
since we add matrices component wise, this looks ok.
/nasser
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1500.2 | norm function search | STAR::ABBASI | | Tue Oct 15 1991 09:24 | 3 |
| ref .-1
Nasser, this function is the same as .0 when p=1 . wake up.
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1500.3 | Looks ok to me | BUZON::BELDIN_R | Pull us together, not apart | Tue Oct 15 1991 09:41 | 5 |
| Your .0 formulation is close to one of the standard normalizations
presented in the vector spaces courses I took thirty years ago. I
predict you will get an acceptable grade on your homework.
Dick
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