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 non-scalar 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.