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Homework Help: Vector space

  1. Jan 29, 2009 #1
    i am given 2 subspaces of R4
    W=sp{(a-b,a+2b,a,b)|a,b[tex]\in[/tex]R}
    U=sp{(1,0,1,1)(-6,8,-3,-2)}
    and am asked to find:
    a homogenic system for W- system for a vector (x,y,z,t) belonging to W

    i see the basis for W is : a(1,1,1,0)+b(-1,2,0,1),, i put these vectors into an extended matrix with (x y z t) on the other side, and after a series of elementary operations, i get
    x+t-z=0
    y-z-2t=0

    next i am asked to find
    a basis for W+U and W[tex]\cap[/tex]U

    forW[tex]\cap[/tex]U
    i find a homogenic system for U, and compre it with the system i found for W which comes to
    x+t-z=0
    y-z-2t=0
    y+8z-8t=0
    3t-4z+x=0
    and i come to
    t=1.5z
    z=z
    y=4z
    x=-0.5z
    so for W[tex]\cap[/tex]U i get a basis (-0.5, 4, 1, 1.5)

    for W+U i take the basis of each and check independace of all of them together, in which i get that all4 are independant, therefore the basis for W+U={(1,0,1,1)(-6,8,-3,-2)(1,1,1,0)(-1,2,0,1),}
    if i perform elementary colum operations on them i can get to (1000)(0100)(0010)(0001), doesnt this mean that W+U is the whole vector space R4 ??

    the final question is
    find a vector other than the zero vector which is orthagonal to all the vectors in U+W
    is this possible, since i found that W+U is the whole vector space R4 (supposing i was correct there)??
     
  2. jcsd
  3. Jan 30, 2009 #2

    Defennder

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    Homework Helper

    Yes it does. In fact a certain theorem tells you that if you can find a set of n linearly independent vectors spanning R^n then that set is a basis for R^n. EDIT: I just checked it and I realised that the vectors are in fact not linearly independent. You need to re-check that part.

    EDIT: As above the vectors are not linearly independent.
     
    Last edited: Jan 30, 2009
  4. Jan 30, 2009 #3
    If W [tex]\cap[/tex] U [tex]\neq[/tex] {0} then the set {basis of U} [tex]\cup[/tex] {basis of W} must be linearly dependent.
     
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