Editing Simplex (section)

=== Volume ===
The [[volume]] of an {{mvar|n}}-simplex in {{mvar|n}}-dimensional space with vertices {{math|(''v''<sub>0</sub>, ..., ''v''<sub>''n''</sub>)}} is
: <math>
\mathrm{Volume} = \frac{1}{n!} \left|\det
\begin{pmatrix}
v_1-v_0 && v_2-v_0 && \cdots && v_n-v_0
\end{pmatrix}\right|
</math>

where each column of the {{math|''n'' × ''n''}} [[determinant]] is a [[vector (geometry)|vector]] that points from vertex {{math|''v''{{sub|0}}}} to another vertex {{math|''v''{{sub|''k''}}}}.<ref>A derivation of a very similar formula can be found in {{cite journal | last1 = Stein | first1 = P. | year = 1966 | title = A Note on the Volume of a Simplex | journal = American Mathematical Monthly | volume = 73 | issue = 3 | pages = 299–301 | jstor = 2315353 | doi = 10.2307/2315353 }}</ref> This formula is particularly useful when <math>v_0</math> is the origin.

The expression
: <math>
\mathrm{Volume} = \frac{1}{n!} \det\left[
\begin{pmatrix}
v_1^\text{T}-v_0^\text{T} \\ v_2^\text{T}-v_0^\text{T} \\ \vdots \\ v_n^\text{T}-v_0^\text{T}
\end{pmatrix}
\begin{pmatrix}
v_1-v_0 & v_2-v_0 & \cdots & v_n-v_0
\end{pmatrix}
\right]^{1/2}
</math>
employs a [[Gram determinant]] and works even when the {{mvar|n}}-simplex's vertices are in a Euclidean space with more than {{mvar|n}} dimensions, e.g., a triangle in <math>\mathbf{R}^3</math>. 

A more symmetric way to compute the volume of an {{mvar|n}}-simplex in <math>\mathbf{R}^n</math> is
: <math>
\mathrm{Volume} = {1\over n!} \left|\det
\begin{pmatrix}
v_0 & v_1 & \cdots & v_n \\
1   & 1   & \cdots & 1
\end{pmatrix}\right|.
</math>

Another common way of computing the volume of the simplex is via the [[Cayley–Menger determinant]], which works even when the n-simplex's vertices are in a Euclidean space with more than n dimensions.<ref>{{mathworld|title=Cayley-Menger Determinant |author=Colins, Karen D. }}</ref>

Without the {{math|1/''n''!}} it is the formula for the volume of an {{mvar|n}}-[[parallelepiped#Parallelotope|parallelotope]]. 
This can be understood as follows: Assume that {{mvar|P}} is an {{mvar|n}}-parallelotope constructed on a basis <math>(v_0, e_1, \ldots, e_n)</math> of <math>\mathbf{R}^n</math>.
Given a [[permutation]] <math>\sigma</math> of <math>\{1,2,\ldots, n\}</math>, call a list of vertices <math>v_0,\ v_1, \ldots, v_n</math> a {{mvar|n}}-path if 
: <math>v_1 = v_0 + e_{\sigma(1)},\ v_2 = v_1 + e_{\sigma(2)},\ldots, v_n = v_{n-1}+e_{\sigma(n)}</math> 
(so there are {{math|''n''!}} {{mvar|n}}-paths and <math>v_n</math> does not depend on the permutation). The following assertions hold:

If {{mvar|P}} is the unit {{mvar|n}}-hypercube, then the union of the {{mvar|n}}-simplexes formed by the convex hull of each {{mvar|n}}-path is {{mvar|P}}, and these simplexes are congruent and pairwise non-overlapping.<ref>Every {{mvar|n}}-path corresponding to a permutation <math>\scriptstyle \sigma</math> is the image of the {{mvar|n}}-path <math>\scriptstyle v_0,\ v_0+e_1,\ v_0+e_1+e_2,\ldots v_0+e_1+\cdots + e_n</math> by the affine isometry that sends <math>\scriptstyle v_0</math> to <math>\scriptstyle v_0</math>, and whose linear part matches <math>\scriptstyle e_i</math> to <math>\scriptstyle e_{\sigma(i)}</math> for all&nbsp;{{mvar|i}}. hence every two {{mvar|n}}-paths are isometric, and so is their convex hulls; this explains the congruence of the simplexes. To show the other assertions, it suffices to remark that the interior of the simplex determined by the {{mvar|n}}-path <math>\scriptstyle v_0,\ v_0+e_{\sigma(1)},\ v_0+e_{\sigma(1)}+e_{\sigma(2)}\ldots v_0+e_{\sigma(1)}+\cdots + e_{\sigma(n)}</math> is the set of points <math>\scriptstyle v_0 + (x_1+\cdots +x_n) e_{\sigma(1)} + \cdots + (x_{n-1}+x_n) e_{\sigma(n-1)} + x_n e_{\sigma(n)}</math>, with <math>\scriptstyle 0< x_i < 1</math> and <math>\scriptstyle x_1+\cdots + x_n < 1.</math> Hence the components of these points with respect to each corresponding permuted basis are strictly ordered in the decreasing order. That explains why the simplexes are non-overlapping. The fact that the union of the simplexes is the whole unit {{mvar|n}}-hypercube follows as well, replacing the strict inequalities above by "<math>\scriptstyle \leq</math>". The same arguments are also valid for a general parallelotope, except the isometry between the simplexes.</ref> In particular, the volume of such a simplex is
: <math> \frac{\operatorname{Vol}(P)}{n!} = \frac 1 {n!}.</math>

If {{mvar|P}} is a general parallelotope, the same assertions hold except that it is no longer true, in dimension&nbsp;> 2, that the simplexes need to be pairwise congruent; yet their volumes remain equal, because the {{mvar|n}}-parallelotope is the image of the unit {{mvar|n}}-hypercube by the [[linear isomorphism]] that sends the canonical basis of <math>\mathbf{R}^n</math> to <math>e_1,\ldots, e_n</math>. As previously, this implies that the volume of a simplex coming from a {{mvar|n}}-path is:
: <math> \frac{\operatorname{Vol}(P)}{n!} = \frac{\det(e_1, \ldots, e_n)}{n!}.</math>

Conversely, given an {{mvar|n}}-simplex <math>(v_0,\ v_1,\ v_2,\ldots v_n)</math> of <math>\mathbf R^n</math>, it can be supposed that the vectors <math>e_1 = v_1-v_0,\ e_2 = v_2-v_1,\ldots e_n=v_n-v_{n-1}</math> form a basis of <math>\mathbf R^n</math>. Considering the parallelotope constructed from <math>v_0</math> and <math>e_1,\ldots, e_n</math>, one sees that the previous formula is valid for every simplex.

Finally, the formula at the beginning of this section is obtained by observing that 
: <math>\det(v_1-v_0, v_2-v_0,\ldots, v_n-v_0) = \det(v_1-v_0, v_2-v_1,\ldots, v_n-v_{n-1}).</math>

From this formula, it follows immediately that the volume under a standard {{mvar|n}}-simplex (i.e. between the origin and the simplex in {{math|'''R'''<sup>''n''+1</sup>}}) is
: <math>{1 \over (n+1)!}</math>

The volume of a regular {{mvar|n}}-simplex with unit side length is
: <math>\frac{\sqrt{n+1}}{n!\sqrt{2^n}}</math>
as can be seen by multiplying the previous formula by {{math|''x''<sup>''n''+1</sup>}}, to get the volume under the {{mvar|n}}-simplex as a function of its vertex distance {{mvar|x}} from the origin, differentiating with respect to {{mvar|x}}, at <math>x=1/\sqrt{2}</math>&nbsp; (where the {{mvar|n}}-simplex side length is 1), and normalizing by the length <math>dx/\sqrt{n+1}</math> of the increment, <math>(dx/(n+1),\ldots, dx/(n+1))</math>, along the normal vector.