Integrated Encryption Scheme

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

$\left(\sum _{i=1}^n{a}_i{c}_i\right)\left(\sum _{j=1}^n{b}_j{d}_j\right)=\left(\sum _{i=1}^n{a}_i{d}_i\right)\left(\sum _{j=1}^n{b}_j{c}_j\right)+\sum _{1\le i<j\le n}(a_i{b}_j-a_j{b}_i)(c_i{d}_j-c_j{d}_i)$

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 ${ℝ}^n$.

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

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

where φ is the angle between the vectors.

Proof

Expanding the last term,

$\sum _{1\le i<j\le n}(a_i{b}_j-a_j{b}_i)(c_i{d}_j-c_j{d}_i)$

$=\sum _{1\le i<j\le n}(a_i{c}_i{b}_j{d}_j+a_j{c}_j{b}_i{d}_i)+\sum _{i=1}^n{a}_i{c}_i{b}_i{d}_i-\sum _{1\le i<j\le n}(a_i{d}_i{b}_j{c}_j+a_j{d}_j{b}_i{c}_i)-\sum _{i=1}^n{a}_i{d}_i{b}_i{c}_i$

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

$det(AB)=\sum _{\genfrac{}{}{0ex}{}{S\subset \{1,\dots ,n\}}{|S|=m}}det(A_S)det(B_S),$

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

$A=\left(\begin{array}{ccc}{a}_1& \dots & a_n\\ b_1& \dots & b_n\end{array}\right),B=\left(\begin{array}{cc}{c}_1& d_1\\ ⋮& ⋮\\ c_n& d_n\end{array}\right).$