Can someone look this over? I have an exam on wednesday and I need to make sure I have the hang of this.

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Let s be a finite set in a vector space V with the property that every x in V has a unique representation as a linear combination of elements of S.  Show that S is a basis of V.
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S = {s[sub]1[/sub] ... s[sub]n[/sub] }

x is in the span S for every x

Let c be a coefficient vector for some linear combination.

c[sub]1[/sub]*s[sub]1[/sub] + ... c[sub]n[/sub]*s[sub]n[/sub] = x has a unique solution (given)
Consider the matrix A = [s[sub]1[/sub]...s[sub]n[/sub]] 

c[sub]1[/sub]*s[sub]1[/sub] + ... c[sub]n[/sub]*s[sub]n[/sub] = A*c = x

therefore A*c = x has a unique solution => A is an invertable matrix => (Invertable Matrix Theorem): the columns of A are linearly independent.

Since S is linear independent and x is in the span of S, S is a basis for V.

Coefficient vector? That seems like unusual terminology..

Showing that S is a basis of V can be done more simply than by the way you are trying to approach it. 

If every vector x in V has a UNIQUE representation as a linear combination of the elements of S, then no one element of S can be a linear combination of another. 

Assume that s1, s2, ..., sn has linear dependencies.  E.g. si = a*sq, where i and q are arbitary vectors in S.  It is given that every vector in V is a unique combination of the vectors in S.  Let's say
x = s3 + 4*s5 + si.  Then x can also be represented as x=s3 + 4*s6 + a*sq.  But then x is not a unique combination of elements of S.  This is a contradiction, so S must be made of a set of linearly independent vectors.  Clearly, Span{S} is the vector space V, and since the elements of S are linearly independent, S must be the basis for the vector space.

At a brief glance your proof seems ok, but I'd use mine over yours on a test without question.  When going through a proof like yours you have to be careful not to make false assumptions - e.g. is A definitely square?  In this case I think it is.

hmm, interesting.

You'll have to tell me how you did so I can ask you questions if I have any when I take this in the fall.  ;)

[quote author=Maddox link=topic=10851.msg107847#msg107847 date=1112944960]
hmm, interesting.

You'll have to tell me how you did so I can ask you questions if I have any when I take this in the fall. ;)
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I don't know. I skipped the class where we got the exam back.
New exam next week.  It's possible that I might fail this class because my teacher sucks ass.