My Math Forum Definition of linear dependence
 User Name Remember Me? Password

 Linear Algebra Linear Algebra Math Forum

May 26th, 2016, 07:35 AM   #11
Senior Member

Joined: Dec 2012
From: Hong Kong

Posts: 853
Thanks: 311

Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics
Quote:
 Originally Posted by zylo A standard definition is: A collection of vectors X1....Xn is LD iff non-zero scalars ai exist st a1X1+....anXn=0 If 2 or more of them are equal, the set is obviously LD. wiki definition: The vectors in a subset $\displaystyle S=\{\vec v_1,\vec v_2,\dots,\vec v_n\}$ of a vector space V are said to be linearly dependent, if there exist a finite number of distinct vectors $\displaystyle \vec v_1,\vec v_2,\dots,\vec v_k$ in S and scalars $\displaystyle a_1,a_2,\dots,a_k$ , not all zero, such that $\displaystyle a_1\vec v_1+a_2\vec v_2+\cdots+a_k\vec v_k= \vec 0$, By phrasing it in this way, they are ruling out the possibility of LD based only on vectors in the set being equal. For Ex: S=(X,X,X1,X2,X3) where (X,X1,X2,X3) are Lineraly Independent. The standard definition would say S is LD, whereas wiki would say the set is LI. Makes sense. EDIT: The case S=(X,2X,X1,X2,X3) where (X,X1,X2,X3) are LI, is ambiguous.
But S is a set, not an ordered n-tuple (which looks like what you've written) or a multiset... Its members have to be distinct, no?

 May 26th, 2016, 08:14 AM #12 Senior Member   Joined: Jun 2015 From: England Posts: 915 Thanks: 271 qwerty you are quite correct to point out, as Archie has already done, that the sets in vector spaces include a uniqueness requirement. Zylo, of all people, should know this since he only accepts the Zermelo set definition. The real purpose of the wiki definition is the following situation for three variables w,x, y w + x = 4 y - x = 6 -w-y = -10 these are linearly dependent although each only contains two of the three variables. Any two are independent. Take the first two, you need a third equation to pick out the third variable. eg combining the first two gives you an equation between w and y since you can eliminate x. but the third equation will not do since it is simply the linear combination multiplied by -1. Note the key phrase S is a subset, not necessarily the entire set. Thanks from 123qwerty Last edited by studiot; May 26th, 2016 at 08:25 AM.
May 26th, 2016, 09:02 AM   #13
Banned Camp

Joined: Mar 2015
From: New Jersey

Posts: 1,720
Thanks: 125

Quote:
 Originally Posted by 123qwerty But S is a set, not an ordered n-tuple (which looks like what you've written) or a multiset... Its members have to be distinct, no?
I use X is a vector.

Technically yes. That's why "collection" is often used.

There is no problem if you want to discuss a general set of vectors S=(X1, X2,... ). where Xi=Xj is a possibility. You could assume this possibility is ruled out by definition of a set, in which case the wiki definition is redundant. But I think that's just a sophistry. You are not thinking about set theory in elementary linear algebra.

So just think about the wiki definition as ruling out equal vectors ruining an otherwise LI set.

The whole thing becomes foggy if you have X and 2X in the set.

Personally, I would use the std def where the type of question raised here is automatically taken care of by context in a particular case:
2 vectors equal? The "set" is LD. But let's check the LD by not considering one of the equal vectors.

X1=2(X2)? Sane thing. Set is LD but you might want to check it without X.

May 26th, 2016, 09:28 AM   #14
Senior Member

Joined: Jun 2015
From: England

Posts: 915
Thanks: 271

Quote:
 There is no problem if you want to discuss a general set of vectors S=(X1, X2,... ). where Xi=Xj is a possibility. You could assume this possibility is ruled out by definition of a set, in which case the wiki definition is redundant. But I think that's just a sophistry. You are not thinking about set theory in elementary linear algebra.
Unfortunately this is not necessarily true when your 'general set' is infinite.

Infinite sets in linear algebra can lead to some unexpected consequences, for example the linear decomposition of a square wave in fourier series.

This is a good reason to stick with the type of vectors I was offering rather than the most general vector space.

 May 27th, 2016, 02:05 AM #15 Senior Member   Joined: Jun 2015 From: England Posts: 915 Thanks: 271 qwerty, do you understand what sort of vector w, x and y are in my example in post 12? They are not geometric or row or column or any sort of n-tuple. They are algebraic variables. That is maps or functions from R to R The rest of the numbers are scalar constants, also from from R This is an example of what you were asking in your post7
May 27th, 2016, 04:26 AM   #16
Senior Member

Joined: Dec 2012
From: Hong Kong

Posts: 853
Thanks: 311

Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics
Quote:
 Originally Posted by studiot qwerty, do you understand what sort of vector w, x and y are in my example in post 12? They are not geometric or row or column or any sort of n-tuple. They are algebraic variables. That is maps or functions from R to R The rest of the numbers are scalar constants, also from from R This is an example of what you were asking in your post7
Yeah I got it. Not straight away, but after a bit. Thanks!

Last edited by 123qwerty; May 27th, 2016 at 04:35 AM.

 Tags definition, dependence, linear

Search tags for this page

### linearly dependent vector maths forum

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post hyperbola Linear Algebra 1 October 19th, 2015 04:57 AM Luiz Linear Algebra 7 August 29th, 2015 12:54 PM Luiz Linear Algebra 1 August 26th, 2015 09:22 AM Luiz Linear Algebra 1 August 24th, 2015 10:45 AM autumn94 Linear Algebra 1 March 29th, 2015 02:14 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.