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Matrix dimension of a vector in R4

  1. Jul 25, 2012 #1
    1. The problem statement, all variables and given/known data
    Let W be the subspace of R4 defined by W={x:V^TX=0}. Calculate dim(w) where
    V=(1 2 -3 -1)^T
    note: V^T means V Transpose, sorry I don't know how to do transpose sign in here.


    2. Relevant equations



    3. The attempt at a solution
    I tries to do it (1 2 -3 -1)(x1 x2 x3 x4)^T=0
    x1+2(x2)-3(x3)-x4=0
    So does it means dim(w)=1 ?
     
  2. jcsd
  3. Jul 25, 2012 #2

    LCKurtz

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    No. I will use a,b,c,d instead of the subscripted x's to save typing. You have ##a+2b-3c-d=0##. Solving for ##a## gives ##a=-2b+3c+d##. So$$
    \left(\begin{array}{c} a \\ b \\ c \\d \end{array}\right) =
    \left(\begin{array}{c} -2b+3c+d \\ b \\ c \\d \end{array}\right) =
    b\left(\begin{array}{c} -2 \\ 1 \\ 0 \\0 \end{array}\right) +
    c\left(\begin{array}{c} 3 \\ 0 \\ 1 \\0 \end{array}\right) +
    d\left(\begin{array}{c} 1 \\ 0 \\ 0 \\1 \end{array}\right)

    $$
    Does that give you a hint about the dimension?
     
    Last edited: Jul 25, 2012
  4. Jul 25, 2012 #3
    It's 3 right? thx man
     
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