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Math Help - Bases over vector spaces.

  1. #1
    Junior Member
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    Unhappy Bases over vector spaces.

    http://img541.imageshack.us/img541/5...2nov211904.gif
    i have this mental block, it's probably really simple but i'm just not getting it
    is this correct?

    for Q, e1=(-1 0 1 0) and e2=(0 -1 0 1)?
    and for F_3, e1=(1 1 0 0), e2=(2 0 1 0) and e3=(1 0 0 1)
    ?

    thank you
    Last edited by mr fantastic; November 22nd 2010 at 11:12 AM. Reason: Deleted begging for help in title.
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  2. #2
    MHF Contributor

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    Actually, you have done extremely well!

    When I did the problem, I got (1, 0, -1, 0) and (0, 1, 0, -1) but since those are just multiples of your vectors, they are both bases. Yes, your answer is correct.

    I assume that your F_3 is the field of integers modulo 3 so that 2= -1 and the two equations we get from applying the matrix \begin{pmatrix}1 & 2 & 1 & 2 \\ 1 & -1 & 1 & -1\end{pmatrix} to any member, \begin{pmatrix}x \\ y \\ z \\ w\end{pmatrix}, of the kernel, x+ 2y+ z+ 2w= 0 and x- y+ z- w= 0, are exactly the same. Since there is only one condition, the dimension of the kernel is 4- 1= 3 rather than 4- 2= 2 as before. We can solve the second equation for x: x= y- z+ w and so \begin{pmatrix}x \\ y \\ z\\ w\end{pmatrix}= \begin{pmatrix}y- z+ w \\ y \\ z \\ w\end{pmatrix} \begin{pmatrix}y \\ y \\ 0 \\ 0\end{pmatrix}+ \begin{pmatrix}-z \\ 0 \\ z \\ 0\end{pmatrix}+ \begin{pmatrix} w \\ 0 \\ 0 \\ w\end{pmatrix}=  y\begin{pmatrix}1 \\ 1 \\ 0 \\ 0\end{pmatrix}+ z\begin{pmatrix}-1 \\ 0 \\ 1 \\ 0\end{pmatrix}+ w\begin{pmatrix}1 \\ 0 \\ 0 \\ 1\end{pmatrix}.

    Since, in F_3, 2= -1, that is exactly what you have.
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