This is a post about affine or projective geometry. It's also a post about intersections. Like, suppose you take two line segments and you want to ask: do these line segments intersect? And suppose you're a little obsessed with linear algebra. So let's use it if we can.

As I mentioned in my last post, one useful aspect of the affine or projective perspective on geometry is that it turns affine subspaces, like points and lines that don't have to go through the origin, into linear subspaces of one larger dimension which do contain the origin and thus can be reasoned about and manipulated with matrices and the tools of linear algebra.

The standard trick here is to imagine your “ambient” space, like the line, the plane, 3-dimensional space and so on, as sitting inside a space of dimension one higher as the collection of points with last coordinate equal to one. Like, if we’re thinking about the line to begin with, put it inside the Cartesian plane as the subspace $$y = 1$$. an affine subspace of the line, like a point, yields (uniquely) a linear subspace of the plane of dimension one higher. So if points are zero-dimensional, we should be expecting a line through the origin—and indeed, for any point on the line $$y = 1$$, like the point $$(x,1)$$, where $$x$$ could be $$2$$ or $$0$$ or $$1,500,743$$, there exists a unique line through the origin and that point. Reasoning about the point and its corresponding line are in some sense the same, provided you eventually look back at the slice $$y = 1$$.

Anyway, suppose we have two affine subspaces of, like, $$\mathbb{R}^n$$ (so think two lines in the plane, for example). Let’s call them $$A$$ and $$B$$. They yield linear subspaces of $$\mathbb{R}^{n+1}$$ as described above, and let’s call these linear subspaces $$A$$ and $$B$$. Now, something like a line segment is usually given to us as two points, right, like the line segment from $$A$$ to $$B$$. Or a plane (or a circle or a triangle in it) might be given to us as three points. The point is that the corresponding vectors in $$\mathbb{R}^{n+1}$$ span the linear subspace. That is, if you have an affine line in $$\mathbb{R}^2$$, say the one through the points $$(3,1)$$ and $$(1,3)$$, the corresponding plane in $$\mathbb{R}^{2+1}$$ is made up of linear combinations of the vectors $$(3,1,1)$$ and $$(1,3,1)$$, and if you look at the collection of those linear combinations with last coordinate equal to $$1$$, you’ll see the line you started with.

Let’s take another line in the plane, like say the line through $$(1,1)$$ and $$(3,3)$$. We know ahead of time that these lines intersect (and in fact the line segments intersect) at the point $$(2,2)$$. But! we can use linear algebra to determine this point of intersection. Abstractly, like, by themselves, without reference to the plane or $2+1$-dimensional space, we have four vectors, so a $4$-dimensional space. We can map these four vectors back into $3$-dimensional space any way we’d like, but let’s say that the line $$A$$ we want to map to positively while the line $$B$$ we’ll map to negatively. Here’s why: if we do so, anything in the intersection of $$A$$ and $$B$$ will be sent to the origin, the vector that is all $$0$$! This mapping we can think of as a matrix and the set of vectors mapped to $$0$$ as the “null space” or solution space of that matrix thought of as a system of linear equations.

More concretely, I’m thinking of the matrix $$\begin{pmatrix} 3 & 1 & -1 & -3 \\ 1 & 3 & -1 & -3 \\ 1 & 1 & -1 & -1 \\ \end{pmatrix}$$. This matrix, you’ll notice, has four columns and three rows. Each of the columns corresponds to one of our points (or really, the vector in $(2+1)$-dimensional space corresponding to that point) and the columns corresponding to the last two points (the second line we chose) are negative. If we think of an abstract combination of the four vectors $$\vec{a}_1$$, $$\vec{a}_2$$, $$\vec{b}_1$$ and $$\vec{b}_2$$, this matrix places those vectors into $(2+1)$-dimensional space by sending $$\vec{a}_1$$ to the vector $$(3,1,1)$$, the vector $$\vec{b}_2$$ to the vector $$(-3,-3,-1)$$ and so on.

One way to find the null space of this matrix is to row-reduce it and then read off from the resulting entries relationships between the corresponding vectors. I won’t explain how to do that; row-reducing is meditative and best practiced “in the privacy of one’s home” as some mathematicians like to say. Anyway, you end up with this matrix $$\begin{pmatrix} 1 & 0 & 0 & -1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & -1 \\ \end{pmatrix}$$. Let’s multiply this matrix by a formal matrix like $$(a_1, a_2, b_1, b_2)$$ and set the result equal to zero. We get $$(a_1 - b_2, a_2 - b_2, b_1 - b_2)$$. For this to be zero, we conclude that $$a_1 = a_2 = b_1 = b_2$$. In other words, an abstract 4-dimensional combination of our points gets mapped to zero by this matrix when it’s an equal combination of the four vectors.

Since we’re not so interested in the abstract points, let’s actually combine our four actual vectors according to this recipe: we add $$(3,1,1)$$ to $$(1,3,1)$$ and $$(1,1,1)$$ and $$(3,3,1)$$ and we get $$(8, 8, 4)$$. This point is, of course, not on the subspace with last coordinate equal to $$1$$, but we can restore that property by dividing through by that last coordinate, the $$4$$. And voila! we get the point $$(2,2,1)$$, which is what we were expecting!

You might be wondering what happens if we take two lines that do not intersect. The curious thing about this process is that the corresponding linear subspaces of $$\mathbb{R}^{2+1}$$ do intersect, (because all linear subspaces of a vector space contain the origin). I encourage you to try it and see!

Just to give you a hint, if you replace the line through $$(3,1)$$ and $$(1,3)$$ with the one through $$(3,1)$$ and $$(5,3)$$ (so this line is parallel to the one through $$(1,1)$$ and $$(3,3)$$), the resulting matrix should be something like $$\begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 1 \\ \end{pmatrix}$$.