Binet–Cauchy identity

In algebra, the Binet–Cauchy identity, named after Jacques Philippe Marie Binet and Augustin-Louis Cauchy, states that

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

for every choice of real or complex numbers (or more generally, elements of a commutative ring). Setting a_i = c_i and b_j = d_j, it gives the Lagrange's identity, which is a stronger version of the Cauchy–Schwarz inequality for the Euclidean space $\scriptstyle \mathbb {R} ^{n}$.

The Binet–Cauchy identity and exterior algebra

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

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

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

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

which can be written as

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

in the n = 3 case.

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

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

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

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

where φ is the angle between the vectors.

Einstein notation

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

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

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

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

Proof

Expanding the last term,

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

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

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

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

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

Generalization

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

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

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

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