Projective Geometry: line representation in P3 I am trying to understand how lines are represented in P3 with the null-space / span representation as described in [1]. The book defines a $2 \times 4$ matrix
$$W = \left[
\begin{matrix}
\mathbf{A}^T \\
\mathbf{B}^T
\end{matrix}
\right], $$
where $\mathbf{A}$ and $\mathbf{B}$ are two distinct points on the line. It further states:
(i) The span of $W^T$ is the pencil of points $\lambda \mathbf{A} + \mu \mathbf{B}$ on the line.
(ii) The span of the 2-dimensional right null space of W is the pencil of planes with the line as axis.
I understand part (ii) as follows: a plane in P3 is defined by a vector $\pi$ of four elements. Point $\mathbf{A}$ lies on a plane if $\mathbf{A}^T \pi = 0$. Same for point $\mathbf{B}$. If $W \pi = 0$ this means that both points lie on $\pi$. There is an infinite number of solutions, which define the pencil of planes which have the line through $\mathbf{A}$ and $\mathbf{B}$ as a common axis. To me, it seems it would be enough to write "The null space of W is the pencil of planes with the line as axis", since the null space are all planes which go through these two points.
Question 1: Why do they state "The span of the 2-dimensional right null space..."?
Part (i) I don't understand at all, since to me the span of $\mathbf{A}$ and $\mathbf{B}$ is a plane and not a line.
Question 2: Can anyone explain part (i) to me?
Thanks a lot!
[1] "Multiple View Geometry in Computer Vision", Hartley and Zisserman, pages 69 and 70
 A: Question 1:
The null space of $W$ is, as you already have found, the set of vectors $\pi$ such that $W\pi = 0.$
Defining this null space by matrix multiplication in this way, since $\pi$ is on the right side of $W$ in the product this vector space is sometimes called the right null space of $W$.
The null space (or right null space) of $W$ is two-dimensional, meaning that there exist two vectors $\pi_1$ and $\pi_2$ (representing two of the planes of the pencil) such that for any plane in the pencil, the vector $\pi$ representing that plane is a linear combination of $\pi_1$ and $\pi_2$.
So the null space of $W$ can also correctly be described (under this construction)
as the two-dimensional right null space of $W.$
The span of any vector space is the vector space itself. So the span of the two-dimensional right null space of $W$ is again the two-dimensional right null space of $W.$ One might ask why we should bother to use the word "span" under these circumstances, since it seems to add nothing to the description. Perhaps the authors had the idea that describing every subspace of each vector space as a "span" of something would be the least confusing way to identify the subspaces. (I'm not saying you have to agree with this idea. I don't think I agree with it. I'm just trying to read the authors' minds to guess why they wrote it the way they wrote it.)

Question 2:
Technically, a point does not have a single set of coordinates in homogeneous coordinates. The coordinates $\begin{bmatrix}1 & \sqrt3 & 2 & 5\end{bmatrix}^T$
and the coordinates $\begin{bmatrix}2 & 2\sqrt3 & 4 & 10\end{bmatrix}^T$
both are homogeneous coordinates of the same projective point.
So saying that $W$ is a $2 \times 4$ matrix whose rows are $\mathbf A^T$ and
$\mathbf B^T$ implies to me that $\mathbf A$ and $\mathbf B$ are not literally
two points of the projective space but are rather merely two representative sets of coordinates of the two desired points, selected either arbitrarily or according to some convention.
The "span" of a matrix seems like a somewhat ambiguous term to me.
In context, I think the "span" of $W^T$ refers to the column space of $W^T$;
that is, we treat each column of $W^T$ as a vector and then take the span of those vectors.
In this case the columns of $W^T$ are literally $\mathbf A$ and $\mathbf B.$
If we regard $\mathbf A$ and $\mathbf B$ as vectors in $\mathbb R^4$,
they span a subspace of $\mathbb R^4$, and every vector in that subspace can be represented as
$$ \lambda \mathbf{A} + \mu \mathbf{B} $$
for some real coefficients $\lambda$ and $\mu.$
This may look like a plane to you, and in fact in $\mathbb R^4$ (viewed as a four-dimensional vector space) this is a plane.
In particular, it is a plane through the origin.
But remember that when we use the four-dimensional real vector space $\mathbb R^4$
as a model of the three-dimensional real projective space $\mathbb{RP}^3,$
each line through the origin of $\mathbb R^4$ models a single point of
$\mathbb{RP}^3$ and each plane through the origin of $\mathbb R^4$
models a single line of $\mathbb{RP}^3.$
So the formula $\lambda \mathbf{A} + \mu \mathbf{B}$
indeed gives you the homogeneous coordinates of a projective line:
we use one degree of freedom to move up and down the line, and the other degree of freedom to generate all the infinitely many equivalent homogeneous coordinates of each projective point we pass through.
