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Topic: Spanning Sets (Read 711 times) |
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Samantha
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Which of the following vectors belong to the subspace of R4 spanned by the set
{(1,1,0,0),(1,-1,0,1),(0,1,1,0)}?
a) (0,0,0,0)
b) (1,0,0,0)
c) (1,6,3,-1)
plz tell me how u arrive at the answer as well
thankyoU!
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Icarus
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Re: Spanning Sets
« Reply #1 on: May 7th, 2004, 3:59pm » |
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The span (smallest subspace containing) of a set of vectors {ri} consists of all vectors v of the form v = [sum] airi for some real numbers ai.
So, v is an element of the span of your three vectors if you can find x, y, z such that
v = x(1,1,0,0) + y(1,-1,0,1) + z(0,1,1,0) = (x+y, x-y+z, z, y).
So, the questions are, can you find x, y, z that give v = (0,0,0,0) ?
Can you find x, y, z that give v = (1,0,0,0)?
Can you find x, y, z that give v = (1,6,3,-1)?
The ones for which the answer is "yes" are in the span. The others are not.
One other hint: Regardless of what your vectors being spanned were, the answer to (a) would have been "yes". 0 is an element of all vector spaces.
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| « Last Edit: May 10th, 2004, 6:14pm by Icarus » |
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Sir Col
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Re: Spanning Sets
« Reply #2 on: May 10th, 2004, 9:49am » |
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Hi Samantha,
Just to amplify what Icarus has already said... every element of a vector space is defined by a minimal set of vectors called a span. For example, a span of [bbr]4 is {(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)}.
For any given vector space,
(i) if a and b belong to the vector space then c=a+b will also be an element.
(ii) if a is an element of the vector space then ra (where r is a real number) will also be an element.
So if {(1,1,0,0), (1,-1,0,1), (0,1,1,0)} is a given span of a subspace of [bbr]4, then they are elements of a vector space, and the results above apply; that is, any multiple and/or linear combination of those vectors also belong to the subspace.
Let u be a vector belonging to the subspace defined by your given span, u=p(1,1,0,0)+q(1,-1,0,1)+r(0,1,1,0)=(p+q,p-q+r,r,q)
If you want to check if a particular vector, (a,b,c,d), belongs to the subspace, write (p+q,p-q+r,r,q)=(a,b,c,d). This gives four linear equations in three unknowns: p+q=a, p-q+r=b, r=c, and q=d.
Taking the third example, by writing, (1,6,3,-1)=(p+q,p-q+r,r,q), we deduce that r=3 and q=-1. Therefore p+q=p-1=1, and we get p=2.
Checking, (p+q,p-q+r,r,q)=(1,6,3,-1), so the vector belongs to the subspace.
I hope that helps.
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| « Last Edit: May 10th, 2004, 2:36pm by Sir Col » |
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Icarus
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Re: Spanning Sets
« Reply #3 on: May 10th, 2004, 4:29pm » |
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We have some terminology differences here. What you are calling a span, I call a basis. In my terminology, which I used in my previous post, the span of any set of vectors is the smallest complete vector space containing the set. This is something different.
I am not saying that one terminology is right and the other wrong, but I want to make clear that we are using the same word for two different things! Otherwise our posts will engender confusion.
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Sir Col
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Re: Spanning Sets
« Reply #4 on: May 10th, 2004, 5:14pm » |
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Hmm? Either I've not made myself clear, or I am misunderstanding the terminology. I was under the impression that a span is a subset of a vector space that is capable of describing every vector in that space; it is sometimes called a generating set. The example I gave for [bbr]4 is a spanning set, but also happens to be its basis, as it describes every possible vector in [bbr]4. The spanning set {(1,1,0,0),(1,-1,0,1),(0,1,1,0)} describes a subspace of [bbr]4, and so it is not a basis.
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Icarus
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Re: Spanning Sets
« Reply #5 on: May 10th, 2004, 6:13pm » |
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Well, clearly you are misunderstanding or mistating something because your definitions are a little mixed up:
Quote:| a span is a subset of a vector space that is capable of describing every vector in that space; it is sometimes called a generating set. The example I gave for [bbr]4 is a spanning set, but also happens to be its basis, as it describes every possible vector in [bbr]4. The spanning set {(1,1,0,0),(1,-1,0,1),(0,1,1,0)} describes a subspace of [bbr]4, and so it is not a basis. |
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So which is it? Does a spanning set have to be capable of describing every vector in that space or not?
Also, note that in my terminology, the span is the vector space, not the spanning set. I have used such terminology as "spanning set" and "this set spans the vector space" as well, but "span" as a noun was reserved for the vector space, and "spanning set" for the subset (which need not be minimal - if it is minimal, then it is a basis for the span).
And I did not say your terminology is wrong. Just that it is different from the one I used (and learned). I am not familiar enough with the body of literature on linear algebra to say that one terminology is standard and the other isn't.
I just wanted to point out the difference in terminology, so that someone less familiar, such as Samantha, would not try to reconcile your definition with mine, and thereby become even more confused since they represent two different things.
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Sir Col
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Re: Spanning Sets
« Reply #6 on: May 11th, 2004, 12:52am » |
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I am certainly no expert in this area, in fact quite the contrary; it is an area that I am at best rusty.
Your explanation about a span not needing to be minimal makes sense, as the definition of a basis is a span that is linear independent, whereas it a spanning set need not be.
I found the following reference in Wikipedia which could well have been written by you. 
http://en.wikipedia.org/wiki/Linear_span
As ever Icarus, I am always grateful for your instruction. Thanks for taking the time to correct me.
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Rob Hunter
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I have this question that i have no idea about. I wonder if any1 can help me?
Find the dimensions of, and basis for, each of the following subspaces of V=R^4 over R:
a) W={(a,b,c,d) are members of V: a+b+c+d=0}
b) U={(a,b,c,d) are members of V: a-b=c-d=0}
c) Y={(a,b,c,d) are members of V: a+b+c=0}
Thanks a lot,
Rob
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towr
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Re: Spanning Sets
« Reply #8 on: Oct 16th, 2004, 1:13pm » |
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Dimensions is easy, just look at the degrees of freedom.
f.i. W has 3 degrees of freedom, if you change the value of any three variables the other is determined (if a=x,b=y,c=z then the only way to get a+b+c+d=0 is to let d=-x-y-z)
This makes the parametrized vector
(a,b,c,d)
= (x,y,z,-x-y-z)
= (x,0,0,-x) + (0,y,0,-y) + (0,0,z,-z)
= x(1,0,0,-1) + y(0,1,0,-1) + z(0,0,1,-1)
which shows a posible basis for the space {(1,0,0,-1), (0,1,0,-1), (0,0,1,-1)}
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