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Algebra, the basis of a solution space

  1. Oct 22, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the basis of the solution space W [itex]\subset[/itex] [itex]\Re^{4}[/itex]
    of the system of linear equations

    [itex]2x_{1}[/itex] + [itex]1x_{2}[/itex] + [itex]2x_{3}[/itex] +[itex]3x_{4}[/itex] =0
    [itex] _{ }[/itex]
    [itex]1x_{1}[/itex] + [itex]1x_{2}[/itex] + [itex]3x_{3}[/itex] = 0


    2. Relevant equations
    The basis must span W and be independent.


    3. The attempt at a solution
    Solving the above system, I get
    [itex]x_{2}[/itex] = [itex]-x_{1}[/itex] - [itex]x_{3}[/itex]
    [itex]x_{4}[/itex] = [itex]\frac{x_{3}-x_{1}}{3}[/itex]

    With 2 degrees of freedom, [itex]x_{1}[/itex] and [itex]x_{3}[/itex],
    so I must need a 2D basis. I separately fixed [itex]x_{1}[/itex] and [itex]x_{3}[/itex] to 1 and the other to zero and got the following vectors:
    [1, -1, 0, -1/3] and [0, -3, 1, 1/3]
    I feel like this is right, as I've been looking up some examples, but I'm not sure this spans all the solutions.

    Am I on the right track?
     
  2. jcsd
  3. Oct 22, 2011 #2

    Mark44

    Staff: Mentor

    Yes, your vectors span W.

    A more systematic way to do things is the row-reduce your matrix, which gives this matrix:
    [tex]\begin{bmatrix}1&0&-1&3\\0&1&4&-3 \end{bmatrix}[/tex]

    From this matrix you can read off your solutions as
    x1 = x3 - 3x4
    x2 = -4x3 + 3x4
    x3 = x3
    x4 = ........ x4

    From this you might be able to see that any vector x in the solution space is a linear combination of these two vectors: <1, -4, 1, 0>T and <-3, 3, 0, 1>T.
     
  4. Oct 22, 2011 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    To add just a little bit, you are saying that for any vector in the solution space, [itex]<x_1, x_2, x_3, x_4>[/itex], we must have [itex]x_2= -x_1- x_3[/itex] and [itex]x_4= (1/3)x_3- (1/3)x_1[/itex]. That is, [itex]<x_1, x_2, x_3, x_4>= <x_1, -x_1- x_3, x_3, (1/3)x_3- (1/3)x_4>= x_1<1, -1, 0, -1/3>+ x_3<0, -1, 1, 1/3>[/itex] which makes it clear what a basis is.

    Mark44 is saying that [itex]x_1= x_3- 3x_4[/itex] and [itex]x_2= -4x_3+ 3x_4[/itex] so that [itex]<x_1, x_2, x_3, x_4>= <x_3- 3x_4, -4x_3+ 3x_4, x_3, x_4>= x_3< 1, -4, 1, 0>+ x_4<-3, 3, 0, 1>[/itex]. That gives another basis for the same subspace.
     
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