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Math Help - Prove that a set is a basis...

  1. #1
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    Prove that a set is a basis...

    Let {v1,v2,v3} be a basis for vector space V. Prove that, if w is not in sp(v1,v2), then S = {v1,v2,w} is also a basis for V.

    I know that in order for S = {v1,v2,w} to be a basis for V, the set S must span V and S must be linearly independent. However, I have no idea whatsoever on how to start this proof. Please help.
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  2. #2
    Senior Member roninpro's Avatar
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    You've identified the necessary items. We can start from there.

    First, suppose that S=\{v_1, v_2, w\} is linearly dependent. Then, we can find c_1,c_2,c_3 (not all zero) so that c_1v_1+c_2v_2+c_3w=0, and we may assume that c_3\ne 0. (Why?) Rearranging gives w=\frac{c_1v_1+c_2v_2}{c_3}. Any problems with this?

    To show \text{span } S=V, we can write w=b_1v_1+b_2v_2+b_3v_3 for some b_1,b_2,b_3, where, in particular, b_3\ne 0. Now, S=\{v_1, v_2, b_1v_1+b_2v_2+b_3v_3\}. Can you see how this helps?

    Good luck.
    Last edited by roninpro; July 12th 2010 at 09:53 PM. Reason: Corrected the placement of "span"
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  3. #3
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    Quote Originally Posted by roninpro View Post
    You've identified the necessary items. We can start from there.

    First, suppose that S=\{v_1, v_2, w\} is linearly dependent. Then, we can find c_1,c_2,c_3 (not all zero) so that c_1v_1+c_2v_2+c_3w=0, and we may assume that c_3\ne 0. (Why?) Rearranging gives w=\frac{c_1v_1+c_2v_2}{c_3}. Any problems with this?

    To show \span S=V, we can write w=b_1v_1+b_2v_2+b_3v_3 for some b_1,b_2,b_3, where, in particular, b_3\ne 0. Now, S=\{v_1, v_2, b_1v_1+b_2v_2+b_3v_3\}. Can you see how this helps?

    Good luck.
    Okay here goes:
    w=\frac{c_1v_1+c_2v_2}{c_3} this basically states that w is dependent of (v1,v2) but since it was given that w is not in span(v1,v2), we can say that c1=c2=c3=0 and S is linearly independent ?

    Span(S) = {rv1 + sv2 + t( b_1v_1+b_2v_2+b_3v_3)} = {c1v1 + c2v2 + c3v3} = Span(V) = V ?

    Is that right?...
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  4. #4
    Senior Member roninpro's Avatar
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    You are correct with linear independence. However, \text{span } S=V doesn't quite follow from what you have written.

    We do have \text{span } S=\{rv_1+sv_2+t(b_1v_1+b_2v_2+b_3v_3)\ | \ r,s,t\in F\}, as you wrote. But this can be rewritten: \{(r+tb_1)v_1+(s+tb_2)v_2+tb_3v_3\ | \ r,s,t\in F\}. To complete the proof, you need to show that r+tb_1, s+tb_2, tb_3 range over all possible values, which amounts to solving a system of equations.
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  5. #5
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    Quote Originally Posted by roninpro View Post
    You are correct with linear independence. However, \text{span } S=V doesn't quite follow from what you have written.

    We do have \text{span } S=\{rv_1+sv_2+t(b_1v_1+b_2v_2+b_3v_3)\ | \ r,s,t\in F\}, as you wrote. But this can be rewritten: \{(r+tb_1)v_1+(s+tb_2)v_2+tb_3v_3\ | \ r,s,t\in F\}. To complete the proof, you need to show that r+tb_1, s+tb_2, tb_3 range over all possible values, which amounts to solving a system of equations.
    Okay... Thanks alot for your help !!
    Also, i just learned in class today that since sp(v1,v2,v3) is a basis for vector space V, we know that dim(V)=3 - which means that any subset of vectors with 3 elements is also a basis for V. I hope that's right.
    Thanks again for your help !!
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  6. #6
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    "...we know that dim(V)=3 - which means that any subset of vectors with 3 elements is also a basis for V..."

    They have to be linearly independent. Then they are a basis.
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  7. #7
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    Quote Originally Posted by Ackbeet View Post
    "...we know that dim(V)=3 - which means that any subset of vectors with 3 elements is also a basis for V..."

    They have to be linearly independent. Then they are a basis.
    ...Yea so i dont have to show that span(S) = V...
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  8. #8
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    BTW... thank you all for your help.. this forum is GREAT !!!
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