I need to prove that $PGL_2(\mathbb{R})\cong SO_3(\mathbb{R})$. Abstract considerations show that both can be identified with the group of projective motions of a conic curve. But maybe there is more explicit isomorphism (in matrix form, for example)?

closed as off topic by Robert Bryant, Misha, Qiaochu Yuan, Angelo, Chris Gerig Apr 23 '13 at 4:30
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Put the bilinear form $\langle, \rangle$ on $2 \times 2$ real matrices by setting $\langle A,B \rangle = {\rm tr}(AB).$ The space of matrices breaks with respect to this form as the orthogonal direct sum of the space of scalar matrices and the $3$dimensional subspace of matrices of trace zero. Now ${\rm GL}(2,\mathbb{R})$ acts by conjugation on the the matrices of trace zero, and preserves this bilinear form in that action. Furthermore, scalar matrices (and nothing more) in ${\rm GL}(2,\mathbb{R})$ are in the kernel of this action, so the action is really one of ${\rm PGL}(2,\mathbb{R}).$ Every matrix in ${\rm GL}(2,\mathbb{R})$ has the eigenvalue $1$ in this action a scalar matrix certainly does and any nonscalar matrix $A$ fixes the matrices of trace zero in ${\rm span}(I,A).$ Every element of ${\rm PGL}(2,\mathbb{R})$ acts with determinant $1$ in this action, as diagonal elements clearly do. This gives an embedding of ${\rm PGL}(2,\mathbb{R})$ in the special orthogonal group determined by this form,and dimension shows that it is surjection. 

