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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

1943.0. "Math Magazine 1467" by RUSURE::EDP (Always mount a scratch monkey.) Thu Feb 23 1995 12:51

    Proposed by John A. Baker, University of Waterloo, Waterloo, Ontario,
    Canada.
    
    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.
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1943.1squares of nth co-ordinates of simplex verticesHERON::BUCHANANEt tout sera bien etFri Mar 03 1995 12:2740
>    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.