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This post gives a general way to calculate $k$-simplex in $n$-dimensional space with $k\leq n$. My question is, if $k=n-1$ and give vertices $v_{0}, \cdots, v_{n-1}$ are linearly independent, can we show that the simplex $S$ generated by $v_{0}, \cdots , v_{n-1}$ has a volume $Vol(S) = \frac{1}{n!}\det\begin{pmatrix}{\bf v}_{0} & \cdots & {\bf v}_{n-1}\end{pmatrix}$?

user124697
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    Surely the volume is zero? The (affine) dimension of the set $\operatorname{aff} { v_0,...,v_{n-1} }$ is $n-1$. For example, if $n=2$ we can take the simplex generated by $e_1,e_2$ which is just s line segment and hence has measure zero. – copper.hat Aug 06 '18 at 02:28
  • @copper.hat Oh, I think the case when we calculate the volume of it as that in the some $n-1$-dimensional hyperplane. – user124697 Aug 06 '18 at 02:34

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I think this is a direct consequence of the Cayley-Menger determinant.

Jimmy Mixco
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