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Linear Independence/Dependence

by Gipson
Tags: linear
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micromass
#37
Dec3-12, 10:06 AM
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Quote Quote by Studiot View Post
I clearly defined the combination of elements and you clearly understood this.

Why are you now asking for a zero set?
The book you quoted clearly talked about addition of two elements. So it makes sense that you would want to define A+B for A and B sets and that you want to have a zero set. If you cannot define these things, then your definition of "linear dependence of sets" is not compatible with the definition of Borowski.

A vector is a set of points that satisfy certain conditions, specific to the problem in hand.
A vector is not a set of points. At least: nobody really thinks of a vector as a set of points. Depending on the set theory you choose, everything is a set. But I doubt many people in linear algebra see (a,b) actually as [itex]\{\{a\},\{a,b\}\}[/itex].

Since this is getting further and further from the OP and personal to boot I withdraw from this thread.
That is perhaps the best decision.
Fredrik
#38
Dec3-12, 01:22 PM
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Quote Quote by Studiot View Post
A vector is a set of points that satisfy certain conditions, specific to the problem in hand.
A vector isn't a set of points. It's just a member of a vector space. A vector space on the other hand can be described as a set of points that satisfy certain conditions, but the only part of those conditions that's "specific to the problem" is the choice of addition operation and scalar multiplication operation. The triple (set,addition,scalar multiplication) must satisfy the usual 8 vector space axioms, and those aren't specific to the problem.
Gipson
#39
Jan26-13, 02:03 PM
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Thx everyone!

Are linearily dependent and independent used properly in the following statements:

1)Inconsistent: Two parallel lines are linearily independent yet have no solutions.

y-x=1
y-x=-2

2)Inconsistent: Three lines intersect at three different points are linearily independent yet have no solutions.

y+2x=-1
y-x=1
y-x=-2

3)Consistent: Three lines intersect at one point are linearily dependent yet have one solution.

y+2x=1
y-x=1
y-2x=1

Given these examples and what I stated in post #1, you can see what my confusion is.


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