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Math Help - Find a basis for this subspace

  1. #1
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    Find a basis for this subspace

    U is a subspace of R^4

    U = {(a,b,c,d) \in R^4| b-2c+d = 0}

    Find a basis for U.

    Unfortunately, I don't even know where to begin. How do I know how many vectors are needed to span U. If someone can help me out and/or start me off, I'd really appreciate it.

    Thanks
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  2. #2
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    Quote Originally Posted by jayshizwiz View Post
    U is a subspace of R^4

    U = {(a,b,c,d) \in R^4| b-2c+d = 0}

    Find a basis for U.

    Unfortunately, I don't even know where to begin. How do I know how many vectors are needed to span U. If someone can help me out and/or start me off, I'd really appreciate it.

    Thanks
    Immediately, you know a=0, and that all 4-tuples that reside in U must then have zero first coordinate.
    Take c and d free. Then b=2c-d. Now you can write, (a,b,c,d) = c(0,2,1,0) + d(0,-1,0,1).
    So dim(U)=2, and a basis for U can be taken as {(0,2,1,0), (0,-1,0,1)}.
    You can now express U by: <{(0,2,1,0), (0,-1,0,1)}>. which just says that U is the set of all linear combinations of the two guys inside the braces.
    Last edited by PiperAlpha167; October 5th 2010 at 11:37 PM.
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  3. #3
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    Thanks for your help but I think you made a mistake.

    a is not 0. a is free and can be anything as it is not mentioned in b-2c+d=0. a can be 100 as long as you have u = (100, 1, 2, 3) or (10,000, 0, 3, 6)...
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  4. #4
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    b- 2c+ d= 0 so d= 2c- b. Every vector (a, b, c, d) is equal to
    (a, b, c, 2c- b)= (a, 0, 0, 0)+ (0, b, 0, -b+ (0, 0, c, 2c)= a(1, 0, 0, 0)+ b(0, 1, 0, -1)+ c(0, 0, 1, 2).
    That tells you everything.
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