Operations for calculating l1 and Frobenius norm

1 Ansicht (letzte 30 Tage)
Xh Du
Xh Du am 12 Jul. 2017
Kommentiert: Xh Du am 13 Jul. 2017
Hi all,
Imagine I have 2 same-length vectors, u1 and u2, and e = u1 - u2. I'd like to calculate 1. l1 norm of e; 2. Frobenius norm of e. In my case I cannot get e and calculate norm directly, so I did something like this for Frobenius norm:
clear; clc;
% test norm 1 and Frobenius norm, what's the equivalent transformation?
u1 = (1:5)';
u2 = u1 - 3;
e = u1 - u2;
%%calculate Frobenius norm directly.
enm2 = norm(e, 'fro');
%%calculate Frobenius norm indirectly
emtx = [u1 -u2];
emtxt = emtx' * emtx;
enmts = sqrt(sum(emtxt(:)));
My question is, can I do something similar for calculating l1 norm? If I get emtx = [u1 -u2], IS there a way to calculate l1 norm without obtaining e = u1 - u2?
Many thanks!

Antworten (1)

Matt J
Matt J am 12 Jul. 2017
Bearbeitet: Matt J am 12 Jul. 2017
Since it's a homework problem, I'll just give you a hint. What does
sum(emtx.^2,2)+2*prod(emtx,2)
give you?
  7 Kommentare
Matt J
Matt J am 13 Jul. 2017
Well, the Frobenius norm and the l1 norm are the same when the dimension of emtx is 1, so whatever you are doing for the Frobenius norm case can be done for the l1 case for each emtx(i). Then you can just add them up...
Xh Du
Xh Du am 13 Jul. 2017
Hi Matt,
I understand that for the above case of 2 vectors, we can do "sum(emtx.^2,2)+2*prod(emtx,2)" to calculate l1 norm. But for more than 2 vectors, we can no longer simply use "sum(emtx.^2,2)+2*prod(emtx,2)", but need:
enm = u1.*u1 + u1.*u2 + u1.*u3 + u2.*u1 + u2.*u2 + u2.*u3 + u3.*u1 + u3.*u2 + u3.*u3
If for 3 vectors, I have
emtx = [u1 u2 u3];
How can I calculate 'enm' from 'emtx' without explicitly writing all the products and sum them?

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