Binet–Cauchy identity

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

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

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 $\mathbb {R} ^{n}$.

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.)