Math 2.0

Geometry Definition

Cayley–Menger determinant

Builds on DeterminantEuclidean distance

Definition

Let n≥1n \ge 1 and N≥1N \ge 1, and let p0,…,pnp_0, \ldots, p_n be points of ℝN\R^N. Write dij=∥pi−pj∥d_{ij} = \norm{p_i - p_j} for their Euclidean distances. The Cayley–Menger determinant of the points is the determinant of the (n+2)×(n+2)(n + 2) \times (n + 2) matrix CM⁡(p0,…,pn)=det⁡(011⋯110d012⋯d0n21d1020⋯d1n2⋮⋮⋮⋱⋮1dn02dn12⋯0).\operatorname{CM}(p_0, \ldots, p_n) = \det \begin{pmatrix} 0 & 1 & 1 & \cdots & 1 \\ 1 & 0 & d_{01}^2 & \cdots & d_{0n}^2 \\ 1 & d_{10}^2 & 0 & \cdots & d_{1n}^2 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & d_{n0}^2 & d_{n1}^2 & \cdots & 0 \end{pmatrix}.

Remarks

The nn-dimensional volume VnV_n of the simplex with vertices p0,…,pnp_0, \ldots, p_n satisfies Vn2=(−1)n+12n(n!)2CM⁡(p0,…,pn).V_n^2 = \frac{(-1)^{n+1}}{2^n (n!)^2} \operatorname{CM}(p_0, \ldots, p_n).

Example: for n=1n = 1 and two points at distance dd, the matrix is (01110d21d20)\begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & d^2 \\ 1 & d^2 & 0 \end{pmatrix}. Its determinant is 2d22d^2, and the formula gives V12=12⋅2d2=d2V_1^2 = \frac{1}{2} \cdot 2d^2 = d^2, the squared length.

Example: for n=2n = 2 and an equilateral triangle of side 11, the matrix is J−IJ - I with JJ the 4×44 \times 4 all-ones matrix. Its eigenvalues are 3,−1,−1,−13, -1, -1, -1, so the determinant is −3-3. The formula gives V22=−116⋅(−3)=316V_2^2 = \frac{-1}{16} \cdot (-3) = \frac{3}{16}, the square of the area 34\frac{\sqrt{3}}{4}.

Degenerate case: if the points are affinely dependent, such as three points on a line, the determinant is 00. The border of ones is essential: the determinant of the matrix of squared distances alone does not give the volume.

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Further reading

  1. Web https://en.wikipedia.org/wiki/Cayley%E2%80%93Menger_determinant
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MSC51K05, 51M25
DefinesCayley–Menger determinant
UsesDeterminant, Euclidean distance

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