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Jun 20, 2023 at 22:22 comment added Bjørn Kjos-Hanssen By the way, important to note that the matrix is $n\times n$ (and not $(n+1)\times (n+1)$) :)
Oct 14, 2010 at 20:08 comment added Suresh Venkat It sounds like this merely reproves the characterization of embeddability in terms of Cayley-Menger determinants ?
Jan 23, 2010 at 23:18 vote accept Matt Noonan
Jan 21, 2010 at 8:25 comment added Andrew Stacey @Mariano: So I understand from the paper. @Tom: as a guess, then, Morgan's contribution was to extend from finite to arbitrary metric spaces. @Everyone else (since this attracted a vote against): My original intention was simply to leave a comment which made Hagen Knaf's answer a little more accessible, but then on reading the paper I decided that the result was simple enough to quote, hence the expansion of a comment into an answer.
Jan 20, 2010 at 15:15 comment added Tom Leinster Andrew: the matrix you mention (let's call it $N$) is closely related to the matrix $M$ in my answer. By an elementary manipulation, your $N$ is positive semidefinite iff my $M$ is conditionally negative semidefinite. In fact, Schoenberg used this equivalence in the 1935 paper that I cited.
Jan 20, 2010 at 12:46 comment added Mariano Suárez-Álvarez Hmm. I guess flat is non-negative determinants?
Jan 20, 2010 at 12:43 history answered Andrew Stacey CC BY-SA 2.5