Binet–Cauchy identity

In algebra, the Binet–Cauchy identity, named after Jacques Philippe Marie Binet and Augustin-Louis Cauchy, states that ^[1]

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

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

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

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

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

When both vectors are unit vectors, we obtain the usual relation

$1=\cos ^{2}(\phi )+\sin ^{2}(\phi )$

where φ is the angle between the vectors.

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

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

In-line notes and references

1. Eric W. Weisstein (2003). "Binet-Cauchy identity". CRC concise encyclopedia of mathematics (2nd ed.). CRC Press. p. 228. ISBN 1-58488-347-2.