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_i = d_i, 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

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

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. (q. e. d.)

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

$\det(AB)=\sum _{S}\det(A_{S})\det(B_{S}),$

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

$A=\left[{\begin{matrix}a_{1}&\dots &a_{n}\\b_{1}&\dots &b_{n}\end{matrix}}\right],\quad B=\left[{\begin{matrix}c_{1}&d_{1}\\\vdots &\vdots \\c_{n}&d_{n}\end{matrix}}\right].$