On page 150 of Richter-Gebert's Perspectives on Projective Geometry, the author mentions that if a conic in $\mathbb{R}\text{P}^2$ of matrix $A$ contains a line of homogeneous coordinates $l$, then the matrix $A$ must necessarily factor into two linear terms $A=l \cdot g^T$ for a suitable vector $g$. Why is that so?
My thought process goes as follows: if the conic contains a whole line, that means that the matrix of that conic is at most of rank 3-1=2. Does that imply directly the result, and if so, how?
The converse is obvious.
It looks like you’re conflating the matrix $A$ with its associated quadratic form $\mathcal Q_A(p)$. What Richter-Gebert writes is that when the quadratic form $\mathcal Q_A(p)$ factors into linear terms, it can be written as $\mathcal Q_{lg^T}(p)$ for suitable vectors $l$ and $g$, i.e., that $p^Tlg^Tp = p^TAp$ for all $p$. This, however, does not imply that $A=lg^T$: the matrix on the right-hand side has rank one, but $A$ could have rank two. Moreover, when dealing with conics $A$ is usually symmetric, but $lg^T$ generally isn’t.
For example, $$A = \begin{bmatrix}1&0&1\\0&-1&0\\1&0&1\end{bmatrix}$$ is a symmetric rank-two matrix that generates the quadratic form $$\begin{align} \mathcal Q_A(p) &= p_1^2-p_2^2+p_3^2+2p_1p_3 \\ &= (p_1+p_2+p_3)(p_1-p_2+p_3) \\ &= p^T\begin{bmatrix}1\\1\\1\end{bmatrix} \begin{bmatrix}1&-1&1\end{bmatrix}p.\end{align}$$ All of the columns of $lg^T$ are scalar multiples of $l$. With $l=(1,1,1)^T$, as shown, no scalar multiple of it can equal any column of $A$. Using the other linear term instead so that $l=(1,-1,1)^T$ doesn’t improve the situation. Note, however, that the symmetric part of $lg^T$, namely, $\frac12(lg^T+gl^T)$ does equal $A$ here. This is to be expected, since only the symmetric part of a matrix contributes to its associated quadratic form.
This factorization is the key to the method described later in the book for splitting a degenerate conic. The author develops an algorithm for finding a skew-symmetric matrix $M$ such that $A+M=lg^T$.