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Math Help - linear algebra

  1. #1
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    linear algebra

    find the basis of annihilator of subspace W in R5, which is spanned by (2,-2,3,4,1), (0,0,-1,-2,3), (-1,1,2,5,2), (1,-1,2,3,0)?

    please explain

    i have reduced to the echo-matrix then i don't have the idea.
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  2. #2
    Senior Member DeMath's Avatar
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    Re: linear algebra

    Use the Gauss–Jordan elimination

    1) reverse rows 1 and 3;
    2) add row 1 to row 3, multiply row 1 by –2 and add row 4;
    3) multiply row 2: by 4 and add to row 3, by –1 and add row 3 to row 1;
    4) devide row 3 by 14 and row 4 by –2;
    5) subtract row 3 from row 4;
    6) multiply row 3 by –3 and add to row 2;
    7) multiply row 2 by 2 and add to row 1



    So, (1,1,0,0,0) and (1,0,-2,1,0).
    Last edited by DeMath; June 12th 2012 at 06:26 PM.
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  3. #3
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    Re: linear algebra

    by inspecting the rank of the matrix DeMath has produced, we see that dim(W) = 3, hence dim(ann(W)) = 2. the vectors DeMath has produced aren't elements of (R5)*, unless you interpret them as the coordinates of the basis elements of ann(W) in the dual basis to the standard basis of R5.

    that is to say, there is a natural isomorphism between ann(W) and W:

    fx given by fx(y) = <x,y> is in ann(W) iff x is in W.

    the chief difference is: ann(W) is a subspace of the dual space, whereas W is a subspace of the original vector space.
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    Re: linear algebra

    by the equations
    x1=x2+x4
    x3=-2x4
    x5=0
    how you are concluding that (1,1,0,0,0) & (1,0,-2,1,0) are in W(per)
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    Re: linear algebra

    Quote Originally Posted by Deveno View Post

    that is to say, there is a natural isomorphism between ann(W) and W:

    fx given by fx(y) = <x,y> is in ann(W) iff x is in W.
    you are saying that from (1,1,0,0,0) and (1,0,-2,1,0), we are able to find fx and fy

    is it right.
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