Content of Cayley-Menger Determinant

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Theorem

Let $V_j$ denote the Cayley-Menger determinant for a $j$-simplex $S$ in $\R^n$.

Let the vertices of $S$ be $v_1, v_2, \ldots, v_{j + 1}$.


Let $B = \sqbrk {\beta_{i j} }$ denote the order $j + 1$ matrix given by:

$\beta_{ij} = {\size {v_i - v_j}_2}^2$

where $\size {v_i - v_j}_2$ is the vector $2$-norm of the vector $v_i - v_j$.


Then the content $V_j$ is given by:

$\map {V_j^2} S = \dfrac {\paren {-1}^{j + 1} } {2^j \paren {j!}^2} \det C $

where $C$ is the order $j + 1$ matrix obtained from $B$ by bordering $B$ with a top row $\tuple {0, 1, \ldots, 1}$ and a left column $\tuple {0, 1, \ldots, 1}^\intercal$.


Proof




Also see


Source of Name

This entry was named for Arthur Cayley and Karl Menger.


Sources