OK, so if we have a a linear transformation $\phi_C$ described with the matrix $C$ in the extended Euclidean plane $E_2^*$ I understand why for finding fixed points we find Eigen-values and we form
$$ (C-\lambda E) \begin{pmatrix} x\\ y\\ t \end{pmatrix}=\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix} $$
But the process for finding coordinates of a fixed line is using the inverse matrix with the coordinates of the line being put before the matrix:
$$ [u_1, u_2, u_3](C^{-1}-\lambda E)=\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix} $$
What's the intuition behind this and why is the multiplication of the coordinates of the line before the matrix and why are we finding the inverse matrix?
Thanks in advance!
The vectors that represent lines are covariant—they don’t transform the same way that points do.
A point $\mathbf p$ is on the line $\mathbf l$ iff $\mathbf l^T\mathbf p=0$. If we have the invertible point transformation $\mathbf p'=C\mathbf p$ then $$\mathbf l^T\mathbf p = \mathbf l^TC^{-1}C\mathbf p = (C^{-T}\mathbf l)^T(C\mathbf p) = (C^{-T}\mathbf l)^T\mathbf p'$$ therefore the line transforms as $\mathbf l'=C^{-T}\mathbf l$. Because of all those transposes, it’s convenient to represent lines as row vectors, in which case the line transformation is $\mathbf l' = \mathbf lC^{-1}$.