Calculating Eigenvalues from two matrices

Let $\alpha$ be the endomorphism given by

$\alpha$:$\mbox{} \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]$$\rightarrow \mbox{} \left[ \begin{array}{cc} d & -b \\ -c & a \end{array} \right] I need to find the eigenvalues and associate eigenspace. Since the determinants are equal, does that mean that the product of the eigenvalues is also the same? And which matrix ought I use to find the eigenvalues? - Do you mean the eigenvalues and eigenspace of the endomorphism itself? If so then the 2x2 identity matrix is an eigenvector with eigenvalue 1. – binn Sep 26 '12 at 2:40 @binn: Yes, of the endomorphism itself. Apologies, but I don't see how its an identity matrix since the Im(\alpha has moved elements inside of the matrix. Sorry if that's confusing... – Edgar Aroutiounian Sep 26 '12 at 2:45 2 Answers It is easy to see that \alpha^2 = I, from which it follows that (\alpha -I)(\alpha + I) = 0. Hence the set of eigenvalues is \{\pm 1 \}. Choose the basis e_1 = \pmatrix{ 1 && 0 \\ 0 && 0 }, e_2 = \pmatrix{ 0 && 1 \\ 0 && 0 }, e_3 = \pmatrix{ 0 && 0 \\ 1 && 0 }, e_4 = \pmatrix{ 0 && 0 \\ 0 && 1 }. In this basis, \alpha has the form A = \pmatrix{ 0 && 0 && 0 && 1 \\ 0 && -1 && 0 && 0 \\0 && 0 && -1 && 0 \\ 1 && 0 && 0 && 0 }. The characteristic polynomial is easily computed to be \det (\lambda I -A) = (\lambda+1)^3 (\lambda-1). Also from A we have \alpha e_2 = -e_2, \alpha e_3 = - e_3, \alpha(e_1+e_4) = e_1+e_4 and \alpha(e_1-e_4) = -(e_1-e_4), which gives all the eigenvectors. - maybe relevant$$ \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