In this article, I introduce the barycentric coordinates: it is an elegant way to represent geometric shapes related to a simplex. By using it, given a simplex, we can construct a hyperellipsoid with the properties: its surface passes every vertex of the simplex, and its tangent hyperplane at each vertex is parallel to the hyperplane containing all other vertices.
and substitute Formula 5 into 4, and then we can derive the quadric of r1
r1TQr1=1.
Note that besides r, there is a 1 in r1, so the quadric is a 2nd-degree polynomial of r, including quadratic terms, linear terms and a constant term.
In order to show that the center of the quadric is a hyperellipsoid whose center is 0, we need to prove that the coefficients of the linear terms are all 0, and the determinant of the coefficients is positive.
Proving that the center of the quadric is 0
Note that Q=(V1V1T)−1, so
Q−1=1V⋯1VT1⋮1=VVT1TVTV1n.
(6)
Substitute Formula 2 into 6, and then we can derive that
Q=0VVT⋯00⋮0n−1=0W⋯00⋮0n1,
where W:=(VVT)−1, so
r1TQr1=rTWr+n1.
The linear terms are all 0, so the center of the quadric is 0.
Proving that the quadric is a hyperellipsoid
We need to show that determinant of the coefficients matrix is positive.
Because Q=(V1−1)TV1−1, we have
∣Q∣=V1−12>0.
Proving that the its tangent hyperplane at vj is parallel to Pj
Here Pj is defined as the hyperplane that passes all vk that k=j.
The equation of the quadric is F(r)=0, where the quadratic function
F(r):=rTWr+n1−1.
According to geometry, the normal vector of the quadric at vj is the gradient of F at vj, which is
νj:=∂r∂F(r)r=vj=2Wvj.
Now consider the normal vector mj of Pj. Assume that
Pj:nmjTr+2=0.
The equation of Pj should holds when r=vk for all k=j, so we can derive n−1 linear equations with respect to mj
∀k=j:nmjTvk+2=0.
(7)
If we can show that
mj=νj=2Wvj
(8)
is the solution to Formula 7, then we can say that the two hyperplane are parallel. Thus, we need to verify the equations derived from substituting Formula 8 into 7
∀k=j:nvjTWvk+1=0,
which is to say that the n×n matrix
P:=VTWV=VT(VVT)−1V
is such a matrix that all of its components except those on its diagonal are −n1.
According to conclusions in matrix analysis, if we regard VT as n−1n-dimensional vectors, then P is an orthogonal projection in Rn to the linear subspace whose basis is the n−1 vectors.
Note that with Formula 2, we can say that the subspace is just a hyperplane whose normal vector is 1. With the conclusion, we can easily write out the form of P because we just need to write out one set of its basis B. Writing out B only requires finding out n−1 linearly independent vectors that are perpendicular to 1. For example,