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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)

IntegrateThis
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  • 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. – aschepler Mar 15 '21 at 16:59

2 Answers2

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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.

Somos
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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.

IntegrateThis
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