Binet–Cauchy identity

When n = 3, the first and second terms on the right hand side become the squared magnitudes of dot and cross products respectively; in n dimensions these become the magnitudes of the dot and wedge products. We may write it

${\displaystyle (a\cdot c)(b\cdot d)=(a\cdot d)(b\cdot c)+(a\wedge b)\cdot (c\wedge d)}$

where a, b, c, and d are vectors. It may also be written as a formula giving the dot product of two wedge products, as

${\displaystyle (a\wedge b)\cdot (c\wedge d)=(a\cdot c)(b\cdot d)-(a\cdot d)(b\cdot c)\,,}$

which can be written as

${\displaystyle (a\times b)\cdot (c\times d)=(a\cdot c)(b\cdot d)-(a\cdot d)(b\cdot c)}$

in the n = 3 case.

In the special case a = c and b = d, the formula yields

${\displaystyle |a\wedge b|^{2}=|a|^{2}|b|^{2}-|a\cdot b|^{2}.}$

When both a and b are unit vectors, we obtain the usual relation

${\displaystyle \sin ^{2}\phi =1-\cos ^{2}\phi }$

where φ is the angle between the vectors.

A relationship between the Levi–Cevita symbols and the generalized Kronecker delta is

${\displaystyle {\frac {1}{k!}}\varepsilon ^{\lambda _{1}\cdots \lambda _{k}\mu _{k+1}\cdots \mu _{n}}\varepsilon _{\lambda _{1}\cdots \lambda _{k}\nu _{k+1}\cdots \nu _{n}}=\delta _{\nu _{k+1}\cdots \nu _{n}}^{\mu _{k+1}\cdots \mu _{n}}\,.}$

The ${\displaystyle (a\wedge b)\cdot (c\wedge d)=(a\cdot c)(b\cdot d)-(a\cdot d)(b\cdot c)}$ form of the Binet–Cauchy identity can be written as

${\displaystyle {\frac {1}{(n-2)!}}\left(\varepsilon ^{\mu _{1}\cdots \mu _{n-2}\alpha \beta }~a_{\alpha }~b_{\beta }\right)\left(\varepsilon _{\mu _{1}\cdots \mu _{n-2}\gamma \delta }~c^{\gamma }~d^{\delta }\right)=\delta _{\gamma \delta }^{\alpha \beta }~a_{\alpha }~b_{\beta }~c^{\gamma }~d^{\delta }\,.}$

Expanding the last term,

${\displaystyle \sum _{1\leq i<j\leq n}(a_{i}b_{j}-a_{j}b_{i})(c_{i}d_{j}-c_{j}d_{i})}$

${\displaystyle =\sum _{1\leq i<j\leq n}(a_{i}c_{i}b_{j}d_{j}+a_{j}c_{j}b_{i}d_{i})+\sum _{i=1}^{n}a_{i}c_{i}b_{i}d_{i}-\sum _{1\leq i<j\leq n}(a_{i}d_{i}b_{j}c_{j}+a_{j}d_{j}b_{i}c_{i})-\sum _{i=1}^{n}a_{i}d_{i}b_{i}c_{i}}$

where the second and fourth terms are the same and artificially added to complete the sums as follows:

${\displaystyle =\sum _{i=1}^{n}\sum _{j=1}^{n}a_{i}c_{i}b_{j}d_{j}-\sum _{i=1}^{n}\sum _{j=1}^{n}a_{i}d_{i}b_{j}c_{j}.}$

This completes the proof after factoring out the terms indexed by i.

A general form, also known as the Cauchy–Binet formula, states the following: Suppose A is an m×n matrix and B is an n×m matrix. If S is a subset of {1, ..., n} with m elements, we write A_S for the m×m matrix whose columns are those columns of A that have indices from S. Similarly, we write B_S for the m×m matrix whose rows are those rows of B that have indices from S. Then the determinant of the matrix product of A and B satisfies the identity

${\displaystyle \det(AB)=\sum _{\scriptstyle S\subset \{1,\ldots ,n\} \atop \scriptstyle |S|=m}\det(A_{S})\det(B_{S}),}$

where the sum extends over all possible subsets S of {1, ..., n} with m elements.

We get the original identity as special case by setting

${\displaystyle A={\begin{pmatrix}a_{1}&\dots &a_{n}\\b_{1}&\dots &b_{n}\end{pmatrix}},\quad B={\begin{pmatrix}c_{1}&d_{1}\\\vdots &\vdots \\c_{n}&d_{n}\end{pmatrix}}.}$