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As it has been done for the Intersection of conics using matrix representation the aim of this page is providing an exaustive and clear numerical example that describe the math behind the decomposition of a degenerate conic. Any help is welcome!


A degenerate conic is defined as a conic whose matrix representation as a vanishing determinant. Geometrically a degenerate conic can be represented by:

  • a pair of real straight lines and an imaginary point
  • a pair of real straight lines and a real point (the intersection of the two lines)
  • a double real point and a pair of imaginary straight lines

The type of degenerate conic depends on the rank of the associated symmetric matrix and the number of real or imaginary eigenvalues.

In practical application it might be necessary to decompose the assiciated matrix $C$ to recover the real stright lines that constitute the degenerate conic.

Typically we have to cover the case of a rank 1 matrix and a rank 2 matrix.

A note on quadratic symmetrization

In a quadratic form (such as $p^T C p$ ) the matrix representation does not need to be a symmetric matrix. It can be proved, though, that $p^T \cdot (C + C^T)/2 \cdot p$ represents the same quadratic form (more specifically a quadratic form depends only on the diagonal entries and on the sums of the terms $C_{ij} + C_{ji}$).

Since $C$ is an homogeneous representation the same quadratic form can be written as $p^T \cdot (C + C^T) \cdot p$. Notice that $C + C^T$ is a symmetric matrix.

Rank 1 degenerate conic decomposition

A rank 1 degenerate conic is represented by a double straight line and has the form $$C = l^T \cdot l$$ Notice that $C$ has rank 1 since it is the composition of rank 1 matrices and it is symmetric.

To decompose $C$ into $l$ we can simply pick the row $i$ (and the column $j$) of $C$ related to any of the non vanishing element $C_{ij}$: $$l \sim C_i $$ (using the line homogeneous representation we can avoid to scale $l$).

Example

Assume that our rank 1 degenerate conics is: $$C_d = \begin{bmatrix}2 & 0 & 2\\0 & 0 & 0\\2 & 0 & 2\end{bmatrix}$$ By selecting the element $C_{d11}$ we can recover the line: $$ l = \begin{bmatrix}2 & 0 & 2\end{bmatrix} \sim \begin{bmatrix}1 & 0 & 1\end{bmatrix}$$

Notice that $$l^T \cdot l= \begin{bmatrix}4 & 0 & 4\\0 & 0 & 0\\4 & 0 & 4\end{bmatrix} \sim C_d$$

Rank 2 degenerate conic decomposition

In case of a rank 2 matrix with real eigenvalues the degenerate conic is composed by two intersecting stright lines $l$ and $m$ and their intersection point $p$.

This conic has the form:

$$ C = l^T \cdot m + m^T \cdot l $$ (the second term is used to obtain a symmetrix matrix).

Moreover any point on one of the two lines $l$ or $m$ belongs to the conic:

If $p$ is on $l$ then $l \cdot p = 0$ and $$ p^T C p = p^T l^T \cdot m p + p^T m^T \cdot l p = 0 \cdot m p + p^T m^T \cdot 0 = 0$$ similarly this can be proven for a point $q$ on line $m$

To decompose this matrix we are going to recover the intersection point $p$ and subsequently the matrix $l^T \cdot m$.

Recovering the intersection point

To localize $p$ we employ the dual conic $B$ associated to $C$. A dual conic represents the locus of all the tangent straight lines to $C$. In particular it can be demonstrated that this locus is still represented by a quadratic form and its matrix form is given by the adjoint matrix $B$ of $C$. The dual conic of a degenerate conic is composed by a double point representing the intersection point of the original degenerate conic's straight lines $l$ and $m$.

It is easily recovered given $B$ by taking the matrix column $B_j$ related to a non vanishing element $B_{ij}$.

In particular: $$ p = \frac{B_j}{ \sqrt{B_{ij}} }$$

Reducing $C$ to rank 1

The intersection point $p$ can be seen as $p = l \times m $. Moreover given $p$ we can recover matrix $S_{l \times m}$ where $p = l \times m$ and $S_{p}$ is the skew symmetrix matrix associated to the vector $p$.

If we now consider $$C_1 = C + S_{l \times m}$$ we can expand it to obtain:

$$ C_1 = l^T \cdot m + m^T \cdot l + S_{l \times m} $$

By expanding each single element this can be transformed in:

$$ C_1 = 2 \cdot l^T \cdot m $$ which is a rank 1 (non symmetrix) matrix.

We can then perform the rank 1 decomposition to recover $l$ and $m$ given $C_1$. Given a non vanishing value $C_{1ij}$ the two lines can be recovered as: $$ l \sim C_{1i} \quad m \sim C_{1j}^T$$

[example]

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closed as unclear what you're asking by José Carlos Santos, Claude Leibovici, mrp, Namaste, Joel Reyes Noche Jun 26 '17 at 12:24

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ As it stands, this post is not a question. Perhaps you should extract the problem statement from it, and move everything else into your own answer. $\endgroup$ – MvG Apr 17 '14 at 13:43
  • $\begingroup$ @Pierluigi: How does one perform this decomposition algorithmically using computer routines? In your other post the lines $l$ and $m$ very much seem to be obtained "by inspection", but how do you do that rigorously? $\endgroup$ – Victor Liu Aug 19 '15 at 22:17
  • $\begingroup$ Simply pick the row (or column) that contains the highest value $\endgroup$ – Pierluigi Aug 23 '15 at 7:23
  • $\begingroup$ @Pierluigi : I like this format. Rather than a question, maybe a local wiki page? Do you have more of these planned? How much .... help ... do you want? I worked out a complete system for handling conics. Nothing fancy, but I had numerical evaluation in mind. I'd love to build a clean, exhaustive guide to handling conics at the machine. $\endgroup$ – Slumberland May 17 '16 at 14:57

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