Below is a passage from a computer vision paper on projective geometry (emphasis added):
A projection from $P^3$ into $P^2$ is represented by a 3×4 matrix $P$,whereby a point x maps to the point $u = Px$. It will be assumed that P has rank 3. Since P has 4 columns but rank 3, there is a unique vector c such that $Pc = (0, 0, 0)^T$. In other words, the projective transformation is undefined at the point $c$, since $(0, 0, 0)^T$ is not a valid homogeneous 3-vector. The point $c$ will be called the perspective centre of the camera.
NOTE: $P^3$ is 3D projective space and $P^2$ is 2D projective space. All that means is that the 3x4 matrix $P$ takes a 3D homogeneous point $(X,Y,Z,1)$ and produces a 2D projection $(x,y,1)$ of that point. Also, homogeneous in this context means having a non-zero last element.
Can someone please explain the bolded sentence to me? Does this have something to do with null space? Why is it that this vector $c$ must exist?
In general, a $3\times4$ matrix $A\colon\mathbb{R}^4\rightarrow\mathbb{R}^3$ must have a non-trivial nullspace: i.e., there must be at least one nonzero vector $u$ for which $Au=0$. This intuitively follows from the fact that we are going down a dimension, so we have to completely collapse at least one direction. In this situation however, there necessarily is infinitely many such vectors, since by linearity, $A(\lambda u)=\lambda Au=\lambda 0=0$ for all $\lambda\in\mathbb{R}$. The fact that you are in projective space, however, geometrically means we are actually sitting on the unit sphere, and so when we project down a dimension, assuming only one dimension gets collapsed, then there can only be one such unit vector $u$ on the unit sphere such that $Pu=0$.