| > Suppose that v[0], v[1], ..., v[n] are the vertices of a regular
> simplex, S, in R^n centered at the origin. Let
>
> v[i] = (v[i,1], v[i,2], ..., v[i,n]) for 0 <= i <= n.
>
> Prove that, for some c > 0,
>
> sum(v[i,j]^2, i=0..n) = c for all j = 1, 2, ..., n.
Let the matrix V be defined as:
(V[0,1] V[0,2] ... V[0,n])
(V[1,1] V[1,2] ... V[1,n])
( . . . )
( . . . )
( . . . )
(V[n,1] V[n,2] ... V[n,n])
Then we are interested in the value of the diagonal elements of V'.V
(where ' denotes transposition.) But what we know is:
(V.V')[j1,j2] = K^2 if j1=j2
K^2.d if j1<>j2
where K is the distance of each point from the origin, and d is the cos
of the angle the two vertices subtend at the origin.
Now, append an extra column to the left of V, so that U = C|V.
U[i,j] = V[i,j] if j > 0
= K*sqrt(-d) if j=0
Then by construction, U.U' = K^2*(1-d)*I (I is the identity matrix.)
So U'.U = U.U'.
U'.U[0,0] = -n*K^2*d => d = -1/(n-1).
So c = K^2*n/(n-1).
Andrew.
|