By using homogeneous coordinates ${\bf x}=[x, y, 1]^T$, a conic section can be expressed by ${\bf x}^T {\bf C} {\bf x} = 0$ with ${\bf C}$ being a 3x3 symmetric matrix. A degenerate form of the conic section is the type "two intersecting lines". When the two lines are also given in homogeneous coordinates with ${\bf g}^T {\bf x}=0$ and ${\bf h}^T {\bf x}=0$, the conic is computed as $${\bf C} = {\bf g} {\bf h}^T + {\bf h} {\bf g}^T$$
How do I extract ${\bf g}$ and ${\bf h}$ from ${\bf C}$?
Interestingly, there seems to be a nice solution for the problem which I found in this Matlab script - however, without any proof. The conic matrix is decomposed using eigendecomposition with $$ {\bf C} = {\bf Q} {\bf \Lambda} {\bf Q}^-1$$ After that, the minimum and maximum eigenvalues $\lambda_1$ and $\lambda_2$, multiplied with the corresponding eigenvectors ${\bf q}_{\text{min}}$ and ${\bf q}_{\text{max}}$ give $${\bf q}_{\text{min}}'=\sqrt{|\lambda_{\text{min}}|} {\bf q}_{\text{min}} $$ $${\bf q}_{\text{max}}'=\sqrt{|\lambda_{\text{max}}|} {\bf q}_{\text{max}} $$ Scaled (and possibly interchanged) versions of ${\bf g}$ and ${\bf h}$ can then be computed with $${\bf g}' = {\bf q}_{\text{min}}' + {\bf q}_{\text{max}}'$$ $${\bf h}' = {\bf q}_{\text{min}}' - {\bf q}_{\text{max}}'$$
Has someone an idea how to derive or proof the solution?
The matrix $\mathbf C$ defines a family of hyperbolas $\mathbf x^T\mathbf C\mathbf x=k$ with common principal axes and asymptotes. The degenerate member of this family $\mathbf x^T\mathbf C\mathbf x=0$ consists of those asymptotes. The asymptotes of a hyperbola in standard position are $\frac xa\pm \frac yb=0$, so they have direction vectors $$\left(\frac1a,\pm\frac1b\right)=\frac1a(1,0)\pm \frac1b(0,1).$$ This has an obvious generalization to a rotated hyperbola: if $\mathbf v_{\text{maj}}$ and $\mathbf v_{\text{min}}$ are unit vectors that give the major (transverse) and minor axes, respectively, with corresponding half-axis lengths $a$ and $b$, its asymptotes have direction vectors $\frac1a\mathbf v_{\text{maj}}\pm\frac1b\mathbf v_{\text{min}}$. (We don’t really need unit vectors for this, but we must have $\|\mathbf v_{\text{maj}}\|=\|\mathbf v_{\text{min}}\|$.)
The eigenvectors of the matrix $\mathbf C$ are the conic’s principal axes and its eigenvalues are the reciprocal squares of the corresponding half-axis lengths, from which the method you found immediately follows.
There’s another method for decomposing such a matrix that doesn’t involve computing eigenvalue and eigenvectors. The idea is to find a skew-symmetric matrix $\mathbf M$ such that $\mathbf C+\mathbf M$ is a scalar multiple of $\mathbf g\mathbf h^T$. The lines can then be read directly from this rank-one matrix: they are a row and column of $\mathbf C+\mathbf M$ that correspond to a non-zero diagonal entry.