Binet–Cauchy identity

Template:Short description 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 Template:Math and Template:Math, it gives Lagrange's identity, which is a stronger version of the Cauchy–Schwarz inequality for the Euclidean space ${ℝ}^n$. The Binet-Cauchy identity is a special case of the Cauchy–Binet formula for matrix determinants.

The Binet–Cauchy identity and exterior algebra

When Template:Math, the first and second terms on the right hand side become the squared magnitudes of dot and cross products respectively; in Template:Math 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 Template:Math, Template:Math, Template:Math, and Template:Math 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 Template:Math case.

In the special case Template:Math and Template:Math, the formula yields $$|a\wedge b{|}^2=|a{|}^2|b{|}^2-|a\cdot b{|}^2.$$

When both Template:Math and Template:Math are unit vectors, we obtain the usual relation $${\sin }^2\varphi =1-{\cos }^2\varphi$$ where Template:Math is the angle between the vectors.

This is a special case of the Inner product on the exterior algebra of a vector space, which is defined on wedge-decomposable elements as the Gram determinant of their components.

Einstein notation

A relationship between the Levi–Cevita symbols and the generalized Kronecker delta is $$\frac{1}{k!}{\epsilon }^{{\lambda }_1\cdots {\lambda }_k{\mu }_{k+1}\cdots {\mu }_n}{\epsilon }_{{\lambda }_1\cdots {\lambda }_k{\nu }_{k+1}\cdots {\nu }_n}={\delta }_{{\nu }_{k+1}\cdots {\nu }_n}^{{\mu }_{k+1}\cdots {\mu }_n}.$$

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 $$\frac{1}{(n-2)!}\left({\epsilon }^{{\mu }_1\cdots {\mu }_{n-2}\alpha \beta }{a}_{\alpha }{b}_{\beta }\right)\left({\epsilon }_{{\mu }_1\cdots {\mu }_{n-2}\gamma \delta }{c}^{\gamma }{d}^{\delta }\right)={\delta }_{\gamma \delta }^{\alpha \beta }{a}_{\alpha }{b}_{\beta }{c}^{\gamma }{d}^{\delta }.$$

Proof

Expanding the last term, $$\begin{array}{cc}& \sum _{1\le i<j\le n}(a_i{b}_j-a_j{b}_i)(c_i{d}_j-c_j{d}_i)\\ =& \sum _{1\le i<j\le n}(a_i{c}_i{b}_j{d}_j+a_j{c}_j{b}_i{d}_i)+\sum _{i=1}^n{a}_i{c}_i{b}_i{d}_i-\sum _{1\le i<j\le n}(a_i{d}_i{b}_j{c}_j+a_j{d}_j{b}_i{c}_i)-\sum _{i=1}^n{a}_i{d}_i{b}_i{c}_i\end{array}$$ 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 $$A=\left(\begin{array}{ccc}{a}_1& \dots & a_n\\ b_1& \dots & b_n\end{array}\right),B=\left(\begin{array}{cc}{c}_1& d_1\\ ⋮& ⋮\\ c_n& d_n\end{array}\right).$$

References

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