Eigentheory of Transformations between Matrix Spaces

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TranscendArcu
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Homework Statement



Screen_shot_2012_02_28_at_12_38_46_PM.png


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

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

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

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

[itex]T \left| \begin{array}{cc}<br /> 0 &0 \\<br /> 0&1 \end{array} \right| =\left| \begin{array}{cc}<br /> 0 &0 \\<br /> 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.
 
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micromass said:
Yes, try to find the matrix of T with respect to your basis. This will be a 4x4 matrix.

But if the matrix is 4x4, how will I apply it to a 2x2 matrix? The inner matrix dimensions won't agree.
 
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.
 
Okay, then I have for my matrix:
[tex] \left| \begin{array}{cccc}<br /> 1 &0&0&0 \\<br /> 0&0&1&0 \\<br /> 0&1&0&0 \\<br /> 0&0&0&1 \end{array} \right| = Q[/tex]
[tex] det(Q- tI) = det\left| \begin{array}{cccc}<br /> 1-t &0&0&0 \\<br /> 0&-t&1&0 \\<br /> 0&1&-t&0 \\<br /> 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?