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Let $\alpha$ be the endomorphism given by

$\alpha$:$\mbox{} \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]$$\rightarrow$ $\mbox{} \left[ \begin{array}{cc} d & -b \\ -c & a \end{array} \right]$

I need to find the eigenvalues and associate eigenspace. Since the determinants are equal, does that mean that the product of the eigenvalues is also the same? And which matrix ought I use to find the eigenvalues?

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Do you mean the eigenvalues and eigenspace of the endomorphism itself? If so then the 2x2 identity matrix is an eigenvector with eigenvalue 1. –  binn Sep 26 '12 at 2:40
    
@binn: Yes, of the endomorphism itself. Apologies, but I don't see how its an identity matrix since the Im($\alpha$ has moved elements inside of the matrix. Sorry if that's confusing... –  Edgar Aroutiounian Sep 26 '12 at 2:45
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2 Answers 2

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It is easy to see that $\alpha^2 = I$, from which it follows that $(\alpha -I)(\alpha + I) = 0$. Hence the set of eigenvalues is $\{\pm 1 \}$.

Choose the basis $e_1 = \pmatrix{ 1 && 0 \\ 0 && 0 }$, $e_2 = \pmatrix{ 0 && 1 \\ 0 && 0 }$, $e_3 = \pmatrix{ 0 && 0 \\ 1 && 0 }$, $e_4 = \pmatrix{ 0 && 0 \\ 0 && 1 }$. In this basis, $\alpha$ has the form $A = \pmatrix{ 0 && 0 && 0 && 1 \\ 0 && -1 && 0 && 0 \\0 && 0 && -1 && 0 \\ 1 && 0 && 0 && 0 }$.

The characteristic polynomial is easily computed to be $\det (\lambda I -A) = (\lambda+1)^3 (\lambda-1)$.

Also from $A$ we have $\alpha e_2 = -e_2$, $\alpha e_3 = - e_3$, $\alpha(e_1+e_4) = e_1+e_4$ and $\alpha(e_1-e_4) = -(e_1-e_4)$, which gives all the eigenvectors.

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maybe relevant

$$ \pmatrix{ 1 & 0 \\ 0 & 1 } \to (+1) \pmatrix{ 1 & 0 \\ 0 & 1 } $$

$$ \pmatrix{ 0 & 0 \\ 1 & 0 } \to (-1) \pmatrix{ 0 & 0 \\ 1 & 0 } $$

$$ \pmatrix{ 1 & 0 \\ 0 & -1 } \to (-1) \pmatrix{ 1 & 0 \\ 0 & -1 } $$

$$ \pmatrix{ 0 & 1 \\ 0 & 0 } \to (-1) \pmatrix{ 0 & 1 \\ 0 & 0 } $$

I would guess that one of the eigenvalues is 1 which corresponds to the 1-dimensional eigenspace

$$ \pmatrix{ x & 0 \\ 0 & x } $$

and that the other eigenvalue is -1 which corresponds to the 3-dimensional eigenspace

$$ \pmatrix{ y & z \\ w & -y } $$

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I'm still not following you... –  Edgar Aroutiounian Sep 26 '12 at 3:01
    
I don't know what I'm doing but someone should give a real answer soon. –  binn Sep 26 '12 at 3:02
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