My understanding is that it should be zero. For example, see section 2.1 in this paper. However, when I try it out by randomly selecting coordinates for the five points and calculating the Cayley-Menger determinant in MATLAB it isn't zero.
Can someone explain this for me? Does this only hold under certain conditions? Or have I misunderstood the whole thing?
These are the points I tried:
P1 : (10, 32, 8)
P2 : (12, 1, 68)
P3 : (45, 107, 24)
P4 : (87, 36, 12)
P5 : (2, 47, 78)
The determinant value I get for the CM is: -0.1565
EDIT: I thought I should explain my logic a bit more. The CM determinants give a relationship for the volume of a simplex as well. If I had four points in $\mathbb{R}^3$ the volume of the tetrahedron would be zero if all four points were on a plane (essentially meaning that they are embedded in $\mathbb{R}^2$). So a generalization would be that if $n$ points are to be embedded in $\mathbb{R}^{n-2}$ their CM determinant should be zero. This is how I concluded that the CM of five points in $\mathbb{R}^3$ should be zero. The thing is I even try the CM determinant for four coplanar points and I still don't get an absolute zero for the CM determinant (I get something near 1e-5).
EDIT2: I can't seem to get the symbolic solver in Matlab to give me a zero for these Cayley-Menger determinants either, but Wolfram-Alpha does.
The determinant in question is clearly zero from a geometric point of view, but we may also give a formal proof. Put the five row vectors together to form a matrix $P$. Let $d$ be the diagonal of $PP^T$ and let $e=(1,1,1,1,1)^T\in\mathbb R^5$. Then the matrix for which we want to calculate the determinant is $$ A=\pmatrix{0&e^T\\ e&de^T+ed^T-2PP^T}. $$ The rank of $P^T$ is at most 3. Hence there exists a nonzero vector $x$ such that $e^Tx=P^Tx=0$. It follows that $A\pmatrix{-d^Tx\\ x}=0$, i.e. $A$ is singular.