# Degenerate conic composed of two lines

In reading the text Multi-View Geometry for Computer vision, after the notion of a conic is introduced with its corresponding conic coefficient matrix, an example is given on page 32 stating

The conic $$C= lm^T + ml^T$$ is composed of two lines $$l$$ and $$m$$. Points on $$l$$ satisfy $$l^T x = 0$$, and are on the conic since $$x^TCx = (x^T l)(m^Tx) + (x^Tm)(l^Tx) = 0$$. Similarly, points satisfying $$m^Tx = 0$$ also satisfy $$x^TCx=0$$.

I'm not sure what this notation means. I'm guessing that $$C$$ is the sum of two outer products, since otherwise it would be a sum of scalars, and then $$x^TCx$$ would be a scalar multiple of the squared norm of $$x$$.

If it is indeed a sum of outer products, then is it generally true that for an outer product $$l \bigotimes m^T$$ that $$x^T l \bigotimes m^T x = (x^T l)(m^Tx)$$ ? (where the latter expression is a product of dot products)

• Note $(x^T l)$ and $(m^T x)$ are scalars, so $(x^T l)(m^T x)$ is two dot products, then an ordinary scalar-scalar multiplication. Commented Mar 15, 2021 at 16:59

We are given column vectors $$x = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}, \quad l = \begin{pmatrix} l_1 \\ l_2 \\ l_3 \end{pmatrix}, \quad m = \begin{pmatrix} m_1 \\ m_2 \\ m_3 \end{pmatrix}.\tag{1}$$ Note that $$l\,m^T := \begin{pmatrix} l_1m_1 & l_1m_2 & l_1m_3 \\ l_2m_1 & l_2m_2 & l_2m_3 \\l_3m_1 & l_3m_2 & l_3m_3 \end{pmatrix} \tag{2}$$ is a rank $$1$$ matrix.

The equation $$l^T x := l_1x_1 + l_2x_2 + l_3x_3 = 0 \tag{3}$$ is the equation representing points $$\,x\,$$ on the line $$\,l.\,$$

The equation $$m^T x := m_1x_1 + m_2x_2 + m_3x_3 = 0 \tag{4}$$ is the equation representing points $$\,x\,$$ on the line $$\,m.\,$$

Define the matrix $$C := lm^T + ml^T = \begin{pmatrix} 2l_1m_1 & l_1m_2+l_2m_1 & l_1m_3+l_3m_1 \\ l_1m_2+l_2m_1 & 2l_2m_2 & l_2m_3+l_3m_2 \\ l_1m_3+l_3m_1 & l_2m_3+l_3m_2 & 2l_3m_3 \end{pmatrix}. \tag{5}$$

This rank $$2$$ matrix represents a degenerate conic which consists of the two lines $$\,l\,$$ and $$\,m.\,$$ The reason is what you mentioned in your question. We have $$x^TCx = (x^Tl)(m^Tx) + (x^Tm)(l^Tx) = 2(l^Tx)(m^Tx)\tag{6}$$ since $$\, l^Tx = x^Tl \,$$ and $$\, m^Tx = x^Tm \,$$ are both scalars. Their product is zero precisely when either of them are zero and each factor is the equation of a line. Hence, this proves the claim about $$\,C\,$$ representing two lines.

Just as a sanity check...

$$C := lm^T + ml^T = \begin{pmatrix} 2l_1m_1 & l_1m_2+l_2m_1 & l_1m_3+l_3m_1 \\ l_1m_2+l_2m_1 & 2l_2m_2 & l_2m_3+l_3m_2 \\ l_1m_3+l_3m_1 & l_2m_3+l_3m_2 & 2l_3m_3 \end{pmatrix}.$$

Then $$C x = \begin{pmatrix} x_1 (2l_1m_1) + x_2 (l_1m_2+l_2m_1) +x_3(l_1m_3+l_3m_1) \\ x_1(l_1m_2+l_2m_1) + x_2(2l_2m_2) +x_3(l_2m_3+l_3m_2) \\ x_1(l_1m_3+l_3m_1) + x_2(l_2m_3+l_3m_2) +x_3(2l_3m_3) \end{pmatrix}$$.

And $$x^T C x =$$ $$x_1(x_1 (2l_1m_1) + x_2 (l_1m_2+l_2m_1) +x_3(l_1m_3+l_3m_1)) + x_2 (x_1(l_1m_2+l_2m_1) + x_2(2l_2m_2) +x_3(l_2m_3+l_3m_2)) + x_3 (x_1(l_1m_3+l_3m_1) + x_2(l_2m_3+l_3m_2) +x_3(2l_3m_3))$$

Whereas $$(x^Tl)(m^Tx) + (x^Tm)(l^Tx) =$$ $$(x_1l_1+x_2l_2+x_3l_3)(m_1x_1+m_2x_2+m_3x_3)$$ $$+ (x_1m_1 + x_2m_2 + x_3m_3)(l_1x_1 + l_2x_2 + l_3x_3)$$

And matching these term by term I see they are equal.