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Math Help - Basis question

  1. #1
    Junior Member
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    Basis question

    For the given matrix H in M2(R) and for any real number r let Vsubr denote the solution space in M2(R) of the equation


    [xy]
    [zw]H =


    rH[xy]
    [zw]

    Find a basis for Vsub2 given that
    H = [01]
    [-10]
    Okay. So I had some difficulty typing this question so it looks really awkward. But I think you can figure out what I'm doing. H is a 2X2 matrix.
    [xy]
    [zw] is supposed to be a 2x2 matrix also.

    thanks.
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  2. #2
    MHF Contributor

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    I think what you are saying is this: Let V_r be the subspace of 2 by 2 matrices satisfying
    \begin{bmatrix}x & y \\ z & w\end{bmatrix}\begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}= r\begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}\begin{bmatrix}x & y \\ z & w\end{bmatrix}.

    Find a basis for that space.

    Okay, go ahead and do the computation: this says
    \begin{bmatrix}-y & x \\ -w & z\end{bmatrix}= \begin{bmatrix}rz & rw \\ -rx & -rw\end{bmatrix}.

    And that reduces to 4 equations: -y= rz, x= rw, -w= -rx, and z= -ry. Since two involve only y and z and two only x and w, that is really two sets of twe equations. Divide the first equation by the last to get -y/z= -z/y which is equivalent to y^2= z^2. That means we must have either y= z or y= -z. But what about r? If y= z, then the first equation above becomes -y= rz so r must be -1. If y= -z, then r must be 1.
    Divide the second equation the third to get -x/w= -w/x or x^2= w^2. We must have either x= w or x= -w. Again, if x= w then r must be -1 and if x= -w, r= 1.

    That is, r must be either 1 or -1. Then matrices in V_{1} are of the form \begin{bmatrix}x & y \\ -y & -x\end{bmatrix} = x\begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}+ y\begin{bmatrix}0 & 1 \\-1 & 0\end{bmatrix} so a basis for this space is the two matrices \begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix} and \begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}.

    Matrices in V_{-1} are of the form \begin{bmatrix} x & y \\ y & x\end{bmatrix} = x\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}+ y\begin{bmatrix}0 & 1 \\ 1& 0\end{bmatrix} so that a basis for V_{-1} is given by the two matrices \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} and \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}
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  3. #3
    Junior Member
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    Yes, but here's the problem. He wants me to find a basis for Vsub2. But you're saying that's impossible. Maybe there is a problem with his question?
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