# Conics consisting of two points/lines makes them rank 2

While studying conics, I came across this concept and example:

Degenerate conics. If the matrix $$C$$ is not of full rank, then the conic is termed degenerate. Degenerate point conics include two lines (rank 2), and a repeated line (rank 1).

Example. The conic

$$C = \mathbf{l}\mathbf{m}^T + \mathbf{m} \mathbf{l}^T$$

is composed of two lines $$\mathbf{l}$$ and $$\mathbf{m}$$. Points on $$\mathbf{l}$$ satisfy $$\mathbf{l}^T \mathbf{x} = 0$$, and are on the conic since $$\mathbf{x}^T C \mathbf{x} = (\mathbf{x}^T \mathbf{l})(\mathbf{m}^T \mathbf{x}) + (\mathbf{x}^T \mathbf{m})(\mathbf{l}^T \mathbf{x}) = 0$$. Similarly, points satisfying $$\mathbf{m}^T \mathbf{x} = 0$$ also satisfy $$\mathbf{x}^T C \mathbf{x} = 0$$. The matrix $$C$$ is symmetric and has rank 2. The null vector is $$\mathbf{x} = \mathbf{l} \times \mathbf{m}$$ which is the intersection point of $$\mathbf{l}$$ and $$\mathbf{m}$$.

Degenerate line conics include two points (rank 2), and a repeated point (rank 1). For example, the line conic $$C^* = \mathbf{x} \mathbf{y}^T + \mathbf{y} \mathbf{x}^T$$ has rank 2 and consists of lines passing through either of the two points $$\mathbf{x}$$ and $$\mathbf{y}$$. Note that for matrices that are not invertible $$(C^*)^* \not= C$$.

I'm wondering why these conics consisting of two points/lines makes them rank 2 (and why is the repeated point for the latter rank 1)? I'd really appreciate clarification of this example. Thank you.

For the two-point/line degenerate conics, the explanation is already there in the text: “The null vector is $$\mathbf x=\mathbf l\times\mathbf m$$” [emphasis mine]. We can drill down into this statement a bit, though.

What is the dimension of the null space of $$\mathbf l\mathbf m^T+\mathbf m\mathbf l^T$$? Well, $$(\mathbf l\mathbf m^T+\mathbf m\mathbf l^T)\mathbf x = (\mathbf m^T\mathbf x)\mathbf l+(\mathbf l^T\mathbf x)\mathbf m = 0.\tag{*}$$ If $$\mathbf l$$ and $$\mathbf m$$ are linearly independent, in which case they represent distinct lines, (*) implies that $$\mathbf l^T\mathbf x = \mathbf m^T\mathbf x = 0$$, in other words, that $$\mathbf x$$ is orthogonal to both $$\mathbf l$$ and $$\mathbf m$$. These vectors are all elements of $$\mathbb R^3$$, so $$\dim\operatorname{span}\{\mathbf l,\mathbf m\} = 2$$, and the dimension of its orthogonal complement and therefore also the nullity of $$\mathbf l\mathbf m^T+\mathbf m\mathbf l^T$$ is $$1$$. Indeed, the orthogonal complement of the span of $$\mathbf l$$ and $$\mathbf m$$ is spanned by $$\mathbf l\times\mathbf m$$.

On the other hand, if $$\mathbf l$$ and $$\mathbf m$$ are linearly dependent, so that both represent the same line, then $$\mathbf l = c\mathbf m$$ for some $$c\ne0$$, and $$\mathbf l\mathbf m^T+\mathbf m\mathbf l^T$$ is a scalar multiple of $$\mathbf m\mathbf m^T$$. If $$\mathbf m\mathbf m^T\mathbf x=0$$, then we must have $$\mathbf m^T\mathbf x=0$$, so the null space of the matrix consists of all vectors orthogonal to $$\mathbf m$$. This is a two-dimensional space, making the rank of the matrix $$1$$. One can also see this directly: the columns of $$\mathbf m\mathbf m^T$$ are all scalar multiples of $$\mathbf m$$, so its column space is spanned by $$\mathbf m$$—its rank is $$1$$.

• Thank you. How did you conclude that $(\mathbf l\mathbf m^T+\mathbf m\mathbf l^T)\mathbf x = (\mathbf m^T\mathbf x)\mathbf l+(\mathbf l^T\mathbf x)\mathbf m$? If you just distributed and rearranged the values, I'm uncomfortable with this, since matrices are not commutative. Sorry for my ignorance. Feb 27, 2020 at 1:42
• @DomFomello $\mathbf m^T\mathbf x$ and $\mathbf l^T\mathbf x$ are scalars.
– amd
Feb 27, 2020 at 4:43
• But what is the algebraic manipulation of the vectors that allowed you to conclude that $(\mathbf l\mathbf m^T+\mathbf m\mathbf l^T)\mathbf x = (\mathbf m^T\mathbf x)\mathbf l+(\mathbf l^T\mathbf x)\mathbf m$? I was not able to understand what was done here. Feb 27, 2020 at 13:52
• @DomFomello Distribute the $\mathbf x$, reassociate, and then because those two dot products are scalars, move them to the left of each term. It’s the same sort of manipulation that’s involved in the formula for an orthogonal projection matrix or a Householder reflection.
– amd
Feb 27, 2020 at 16:38
• @DomFomello That’s right.
– amd
Feb 28, 2020 at 15:05

I was also really confused by the lack of explanation of some topics in this book. To have a better understanding of degenerate conics in 2D projective space, I had to look for alternative sources of info. Watching a visualization of conics and degenerate conics on youtube helped me out a lot. I finally stumbled upon this pdf:

"In degenerate cases, it may degenerate to two lines when $$\operatorname{rank}(C) = 2$$, or one repeated line when $$\operatorname{rank}(C) = 1$$."
Meaning that the rank is for the matrix $$C$$. Then I found this presentation of the University of Berkeley about Multiple view geometry: