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Thread: Finding a vector space

  1. #1
    Sep 2008

    Finding a vector space

    Hey, I have a question here that asks:

    Find a basis of the vector space U.
    U = {(x1, ..., X6) belongs to R6 | x1 = 2x2, x3 = 7x6}

    I realize the basis has to have four elements and that I have to find a set of generators but I do not know where to go from there. Any help would be appreciated. Thanks.
    Last edited by GreenDay14; Sep 30th 2009 at 08:10 AM. Reason: Forgot a line
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  2. #2
    MHF Contributor

    Apr 2005
    R^6 is, of course, 6 dimensional and, generally, each equation reduces the dimension by 1 so your subspace is 6- 2= 4 dimensional and you need a basis containing 4 vectors.

    Here's a good way of finding that basis. You are given that x1= 2x2 and x3= 7x6. That means that you are not "free" to choose x1 and x3 to be whatever you want- they are fixed by the values of x2 and x6. But all the others are "free". Take those to be, one at a time, equal to 1 while the other "free variables" are 0.
    For example, if x2= 1, x4= x5= x6= 0 (I have skipped x1 and x3 because they are not "free") then x1= 2x2= 2 and x3= 7x6= 0 so we have <2, 1, 0, 0, 0, 0> as one basis vector. If x4= 1, x2= x5= x6, then x1= 2x2= 0 and x3= 7x6= 0 so we have <0, 0, 0, 1, 0, 0> as another basis vector. If x5= 1, x2= x4= x6= 0, then x1= 2x2= 0 and x3= 7x6= 0 so we have <0, 0, 0, 0, 1, 0> as a third basis vector. Finally, if x6= 1, x2= x4= x5= 0, then x1= 2x2= 0 and x3= 7x6= 7 so we have <0, 0, 7, 0, 0, 1> as the fourth basis vector.
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