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Math Help - Is V in Span(S)

  1. #1
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    Is V in Span(S)

    Hey,

    I have a question that I'm not sure how to approach at all. The question is:

    Let

    S={[[-1],[2],[1]] , [[3],[1],[2]] , [[1],[5],[4]] , [[-6],[5],[1]]}

    and

    v=[[-5],[3],[0]]

    Is v in span(S)?

    So i know that a set of vectors ie. v, spans a subspace, S, if every vector in S can be written as a linear combination of vectors in v.

    Thanks in advance for any help you might be able to give!!
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  2. #2
    Senior Member MacstersUndead's Avatar
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    Re: Is V in Span(S)

    Alright, let's us definition as you laid out. If v is a linear combination of vectors in S, then there exists constants c_i such that

    c_1(-1,2,1) + c_2(3,1,2) + c_3(1,5,4) + c_4(-6,5,1) = (-5,3,0)

    by comparison we get the simultaneous system of equations

    -c_1 + 3c_2 + 1c_3 + -6c_4 = -5
    2c_1 + c_2 + 5c_3 + 5c_4 = 3
    1c_1 + 2c_2 + 4c_3 + 1c_4 = 0

    You can solve this system using a matrix. If such c_i exist, then you have show v is in a span of S. If no such c_i exist then v is not in the span of S.
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    Re: Is V in Span(S)

    Quote Originally Posted by MacstersUndead View Post
    Alright, let's us definition as you laid out. If v is a linear combination of vectors in S, then there exists constants c_i such that

    c_1(-1,2,1) + c_2(3,1,2) + c_3(1,5,4) + c_4(-6,5,1) = (-5,3,0)

    by comparison we get the simultaneous system of equations

    -c_1 + 3c_2 + 1c_3 + -6c_4 = -5
    2c_1 + c_2 + 5c_3 + 5c_4 = 3
    1c_1 + 2c_2 + 4c_3 + 1c_4 = 0

    You can solve this system using a matrix. If such c_i exist, then you have show v is in a span of S. If no such c_i exist then v is not in the span of S.
    So when I row reduced that matrix, I ended up with [[1,0,2,3,2],[0,1,1,-1,-1],[0,0,0,0,0]]

    Seeing as this means that there are infinitely many solutions does this mean that v is NOT in span(S)?
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  4. #4
    Senior Member MacstersUndead's Avatar
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    Re: Is V in Span(S)

    No, the solution for c_i need not be unique in order for v to be in the span (choose one linear combination and you're done). In fact, you are bound to have at least one free variable since you have four variables and three equations.
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    Re: Is V in Span(S)

    Ok, so because there is a solution, even if it is infinitely many, we can say that v is in span s? Thank you very much for your time, you have been a great help by the way! So how could v be expressed as a linear combination of the vectors from s then? Am i correct in saying that it would be:

    if we let c_3=t and c_4=s


    (2-2t-3s)(-1,2,1)+(-1-t-s)(3,1,2)+t(1,5,4)+s(-6,5,1)=(-5,3,0)
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  6. #6
    Senior Member MacstersUndead's Avatar
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    Re: Is V in Span(S)

    This looks correct. Well done.
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    Re: Is V in Span(S)

    Thank you so much for your time!! I was dreading trying to figure out how to do this question, you have been such a great help
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  8. #8
    Senior Member MacstersUndead's Avatar
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    Re: Is V in Span(S)

    No problem, I've been there.

    I also forgot to mention the following; if you are comfortable in the rationale in setting up a matrix you can "short-cut" by inserting the spanning vectors into the matrix as columns from left to right (doesn't matter the order you insert the vectors in) with the vector that you want spanned at the end. (set up a couple of times and you'll see why)
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  9. #9
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    Re: Is V in Span(S)

    Thanks I'll be sure to check it out!
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