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Finding the basis of this subspace

  1. Jan 24, 2009 #1
    in linear algebra, if i am told to find a basis for the following

    W={(x,y,z,t)|x+y=0, x+t=0}

    what i did was

    1 1
    1 0
    0 0
    0 1
    after performing elementary actions on rows, i came to

    1 0
    0 1
    0 0
    0 0
    from here i can see that they are linearly independant and they cleary span their own space so they are my basis.

    basis= (1,1,0,0) (0,1,0,1)

    the correct solution in my book was quite different, they put the vectors lying down like so

    1 1 0 0
    0 1 0 1

    and eventually got to

    1 0 0 -1
    0 1 0 1

    what is the reason that my way is wrong,
    Last edited: Jan 24, 2009
  2. jcsd
  3. Jan 24, 2009 #2


    Staff: Mentor

    I think you might have started off with the wrong vectors. Based on your definition of W, I see that
    x = -y
    y = y
    z = z
    t = t

    Since x = -y and x = -t, it's clear that y = t, so the four equations above simplify to:
    x = -t
    y = t
    z = z
    t = t
    This set of equations can be written as
    (x, y, z, t) = t(-1, 1, 0, 1) + z(0, 0, 1, 0)

    So two vectors that span W are the two above, and they are clearly linearly independent, since no multiple of either could possibly produce the other. (It's very easy to check independence when you have only two vectors. With three or more you can't do it by inspection.)

    Notice that these two vectors satisfy the equations that are part of your definition; your vectors don't.
  4. Jan 24, 2009 #3
    Mark's work is absolutely correct. Let's just make sure you know what you're doing with those matrices.

    The span of W is the null space of the matrix

    [tex]A = \begin{pmatrix} 1 & 1 & 0 & 0\\ 1 & 0 & 0 & 1\end{pmatrix}[/tex]

    ,i.e., is the solution set of Av=0 where v=(x,y,z,t).

    By row reducing (just one operation) Av=0 we see that the null space of A is equal to the null space of

    [tex]A' = \begin{pmatrix} 1&1&0&0\\0&-1&0&1\end{pmatrix}[/tex].

    ,i.e., the solution set of A'v = 0.

    With v = (x,y,z,t), writing this out we have x+y=0 and -y+t=0. The solution set of this follows from Mark's work (thus illustrating that matrices in this particular example are quite unnecessary).

    Presuming these matrices are supposed to represent A (from our work above), check you're referring to the same question! Our set-up has x+t=0 whereas the row "0 1 0 1" suggests they have y+t=0 !
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