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Finding a spanning set

  1. Nov 27, 2008 #1
    1. The problem statement, all variables and given/known data

    Let W be a subspace of V = [tex]C^5[/tex] consisting of all vectors [tex]
    x = (x_1,x_2,x_3,x_4,x_5) \epsilon C^5[/tex] that satisfy:

    [tex]
    -2ix_1 + x_2 - x_3 + (1 - i)x_4 = 0
    [/tex]
    [tex]
    x_1 + ix_2 -2x_5 = 0
    [/tex]


    Find a set that spans W.


    2. Relevant equations



    3. The attempt at a solution

    From the second equation we know that [tex] x_1 = 2x_5 - ix_2[/tex]. Substituting that into the first, we have [tex]-2i(2x_5 -ix_2) + x_2 - x_3 + (1-i)x_4 = 0[/tex]. Expanding through the brackets and simplifying, we have [tex]-4ix_5 - x_2 - x_3 + (1-i)x_4 = 0[/tex]. This implies that [tex]x_2 = -x_3 +(1-i)x_4 -4ix_5[/tex]. So that is the only condition we have. So, we now have the following:

    [tex](1, 0, 0, 0, 0)x_1 + (0, -1, 1, 0, 0)x_3 + (0, 1-i, 0, 1, 0)x_4 + (0, -4i, 0, 0, 1)x_5[/tex]


    So the set [tex]{ (1, 0, 0, 0, 0), (0, -1, 1, 0, 0), (0, 1-i, 0, 1, 0), (0, -4i, 0, 0, 1)[/tex] should span the space. However, if you multiply the first vector by 1, and the rest by 0, we have [tex](1,0,0,0,0)[/tex], which obviously doesn't satisfy the two equations above.

    What am I doing wrong?
     
  2. jcsd
  3. Nov 27, 2008 #2

    lurflurf

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    badly worded W is a set that spans W
    assume a minimal spaning set is desired
    start by assuming wlog that the form of the vectors is
    (1,0,0,*,*)
    (0,1,0,*,*)
    (0,0,1,*,*)
    where the * are chosen to satisfy the equations.
     
  4. Nov 27, 2008 #3
    I was wondering if you could show me where I went wrong with the method I was using?
     
  5. Nov 27, 2008 #4

    lurflurf

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    Many things. For a start you have combined the two conditions into one. This is wrong. Sometimes conditions are superfluous and can be discarded, not in this case. It is true that if a vector satisfies both of the given equations it will satisfy your, it is untrue that a vector that satisfies your condition satifies either of those given.
    I sugest you abandon this method
    For example as you yourself point out the vector
    (1,0,0,0,0) satisfies your condition, but not those given.
    Your are free to use your condition, but if you do you need to also a condition such as
    x1=-i*x3+(i-1)x4-2x5
    which none of your proposed vectors satisfy.
    another example say you desired to solve
    x+y=3
    x-y=1
    it is true that the solution satisfies
    x=2
    but (2,11) is not a solution
    in both examples you have introduced an extraneous or spurious solution.
    Extraneous or spurious solutions are solution introduced by manimulation of the given conditions, they are not solutions of the original equation.
     
  6. Nov 27, 2008 #5
    lurflurf, I know that you recommend me to approach the problem differently, but I'm quite adamant on solving it this way (I've solved similar questions this way before so it is really bugging me that I cannot solve this one). Please don't feel as if I'm ignoring you.

    With that said, why is combining the two equations into one equation wrong? All I did was a simple substitution.
     
  7. Nov 27, 2008 #6
    Nevermind..
     
  8. Nov 27, 2008 #7

    lurflurf

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    They are not really combined. You can exchange the two given equations for another two if you like. Sometimes that is very helpful, though in this case not so much. The wrong part is your one equation does not contain all the information of the original two. You can use your equation with another if you like, but it is wrong to use it alone.

    as I said before you could use
    x1=-i*x3+(i-1)*x4-2*x5
    x2=-x3+(1-i)*x4-4i*x5
    with these equation your could work out the * in
    (*,*,1,0,0)
    (*,*,0,1,0)
    (*,*,0,0,1)
     
  9. Nov 27, 2008 #8
    "x1=-i*x3+(i-1)*x4-2*x5"

    How did you get that?

    I'm more concerned with the method of solving this problem.
     
  10. Nov 27, 2008 #9

    lurflurf

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    oops that should be
    x1=i*x3+(i-1)*x4-2*x5

    we have
    -2i*x1+x2-x3+(1-i)*x4=0
    x1+i*x2-2*x5=0

    multiply the first by i and the second by -1 to give
    2*x1+i*x2-i*x3+(-1+i)*x4=0
    -x1-i*x2+2*x5=0

    add the two equations to give
    x1-i*x3+(-1+i)*x4+2*x5=0


    solve for x1 to give
    x1=i*x3+(1-i)*x4-2*x5

    method got it
    Step 1:
    We want a number of vector (which we will later see is 3) that satisfy the conditons and such that every solution to the equations is contained in their span.
    Step 2:
    chose a basis for the space
    step 3:
    take nice vector(s) and adjust it to satisfy the equations
    step 4:
    use adjusted nice vector(s) to span subspace
     
  11. Nov 27, 2008 #10
    [tex] {(i,-1,1,0,0), (1-i, 1-i, 0,1,0), (-2, -4i, 0, 0,1)}[/tex] shall be our spanning set. :)

    Thanks a lot lurflurf! Did you know that we had to represent [tex]x_1[/tex] in terms of [tex]x_3, x_4, x_5[/tex] because I was "loosing" information about it by trying to describe a two equation system of 5 variables in just one equation? Also, how did you know that a minimal set would be three vectors?
     
  12. Nov 27, 2008 #11
    I see now. [tex]x_1[/tex] wasn't being described at all! Thus, you had to describe [tex]x_1[/tex] in terms of the same vectors that [tex]x_2[/tex] was described in, which was the other three vectors, which was how you knew a minimal set would consist of three vectors.

    Thanks!
     
  13. Nov 27, 2008 #12

    lurflurf

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    Yes you were losing information. An equation tends to reduce the dimension by 1. In this problem we started with dimension 5 and had 2 equations so we had 5-2=3 dimensions in the end. It is also poeesible that equations are redundent and add no information. Also equations can be inconsistent and have no solution.
    x+y=3
    x-y=1
    is consistent with 1 solution (x,y)=(2,1)
    2x+2y=4
    3x+3y=6
    is consistent
    the equations are redundent
    x+y=1
    x+y=2
    is inconsistent
    there are no solutions
     
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