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Eigentheory of Transformations between Matrix Spaces

  1. Feb 28, 2012 #1
    1. The problem statement, all variables and given/known data

    Screen_shot_2012_02_28_at_12_38_46_PM.png

    My instructor wants me to only solve for the case m=2.

    3. The attempt at a solution

    So I thought I should discover what T does to the standard basis for matrices of size 2x2:

    [itex]T \left| \begin{array}{cc}
    1 &0 \\
    0&0 \end{array} \right| = \left| \begin{array}{cc}
    1 &0 \\
    0&0 \end{array} \right|[/itex]

    [itex]T \left| \begin{array}{cc}
    0 &1 \\
    0&0 \end{array} \right| =\left| \begin{array}{cc}
    0 &0 \\
    1&0 \end{array} \right|[/itex]

    [itex]T \left| \begin{array}{cc}
    0 &0 \\
    1&0 \end{array} \right| =\left| \begin{array}{cc}
    0 &1 \\
    0&0 \end{array} \right|[/itex]

    [itex]T \left| \begin{array}{cc}
    0 &0 \\
    0&1 \end{array} \right| =\left| \begin{array}{cc}
    0 &0 \\
    0&1 \end{array} \right|[/itex]

    At this point do I have to create a matrix consisting of these transformed matrices in order to find [itex]Δ_T (t) = det(M - tI)[/itex]? I'm kind of confused about my next step.
     
  2. jcsd
  3. Feb 28, 2012 #2

    micromass

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    Yes, try to find the matrix of T with respect to your basis. This will be a 4x4 matrix.
     
  4. Feb 28, 2012 #3
    But if the matrix is 4x4, how will I apply it to a 2x2 matrix? The inner matrix dimensions won't agree.
     
  5. Feb 28, 2012 #4

    micromass

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    The 2x2-matrices won't have anything to do with the matrix of T.

    The entries of the matrix of T are coordinates.

    For example, if we put on [itex]M_{2,2}(\mathbb{R})[/itex] (the 2x2-matrices) the basis you mention, then the 4x4-matrix will contain the coordinates of the image of the basis.

    For example, you have to write

    [tex]T\left(\begin{array}{cc} 1 & 0\\ 0 & 0\end{array}\right) [/tex]

    as a linear combination of the basis. So we have

    [tex]T\left(\begin{array}{cc} 1 & 0\\ 0 & 0\end{array}\right) = 1 \left(\begin{array}{cc} 1 & 0\\ 0 & 0\end{array}\right) + 0\left(\begin{array}{cc} 0 & 1\\ 0 & 0\end{array}\right) + 0\left(\begin{array}{cc} 0 & 0\\ 1 & 0\end{array}\right) +0\left(\begin{array}{cc} 0 & 0\\ 0 & 1\end{array}\right)[/tex]

    So the coordinates of

    [tex]T\left(\begin{array}{cc} 1 & 0\\ 0 & 0\end{array}\right)[/tex]

    are (1,0,0,0). This will be the first column of the 4x4-matrix.
     
  6. Feb 28, 2012 #5
    Okay, then I have for my matrix:
    [tex]
    \left| \begin{array}{cccc}
    1 &0&0&0 \\
    0&0&1&0 \\
    0&1&0&0 \\
    0&0&0&1 \end{array} \right| = Q
    [/tex]
    [tex]
    det(Q- tI) = det\left| \begin{array}{cccc}
    1-t &0&0&0 \\
    0&-t&1&0 \\
    0&1&-t&0 \\
    0&0&0&1-t \end{array} \right| = (1-t)(1-t)[(t)(t) - 1] = (t-1)^3 (t+1)
    [/tex]

    That is the characteristic polynomial, and its roots are the eigenvalues. So we have eigenvalues 1,-1. Look about right?
     
  7. Feb 28, 2012 #6

    micromass

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    That sounds correct. Now you need to find the eigenspaces.
     
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