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

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

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 Lagrange's identity, which is a stronger version of the Cauchy–Schwarz inequality for the Euclidean space ${\textstyle \mathbb {R} ^{n}}$. The Binet-Cauchy identity is a special case of the Cauchy–Binet formula for matrix determinants.

The Binet–Cauchy identity and exterior algebra

When 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

$${\displaystyle (a\cdot c)(b\cdot d)=(a\cdot d)(b\cdot c)+(a\wedge b)\cdot (c\wedge d)}$$

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

$${\displaystyle (a\wedge b)\cdot (c\wedge d)=(a\cdot c)(b\cdot d)-(a\cdot d)(b\cdot c)\,,}$$

which can be written as

$${\displaystyle (a\times b)\cdot (c\times d)=(a\cdot c)(b\cdot d)-(a\cdot d)(b\cdot c)}$$

in the n = 3 case.

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

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

When both a and b are unit vectors, we obtain the usual relation

$${\displaystyle \sin ^{2}\phi =1-\cos ^{2}\phi }$$

where φ 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

$${\displaystyle {\frac {1}{k!}}\varepsilon ^{\lambda _{1}\cdots \lambda _{k}\mu _{k+1}\cdots \mu _{n}}\varepsilon _{\lambda _{1}\cdots \lambda _{k}\nu _{k+1}\cdots \nu _{n}}=\delta _{\nu _{k+1}\cdots \nu _{n}}^{\mu _{k+1}\cdots \mu _{n}}\,.}$$

The ${\displaystyle (a\wedge b)\cdot (c\wedge d)=(a\cdot c)(b\cdot d)-(a\cdot d)(b\cdot c)}$ form of the Binet–Cauchy identity can be written as

$${\displaystyle {\frac {1}{(n-2)!}}\left(\varepsilon ^{\mu _{1}\cdots \mu _{n-2}\alpha \beta }~a_{\alpha }~b_{\beta }\right)\left(\varepsilon _{\mu _{1}\cdots \mu _{n-2}\gamma \delta }~c^{\gamma }~d^{\delta }\right)=\delta _{\gamma \delta }^{\alpha \beta }~a_{\alpha }~b_{\beta }~c^{\gamma }~d^{\delta }\,.}$$

Proof

Expanding the last term,

$${\displaystyle {\begin{aligned}&\sum _{1\leq i<j\leq n}(a_{i}b_{j}-a_{j}b_{i})(c_{i}d_{j}-c_{j}d_{i})\\={}&{}\sum _{1\leq i<j\leq n}(a_{i}c_{i}b_{j}d_{j}+a_{j}c_{j}b_{i}d_{i})+\sum _{i=1}^{n}a_{i}c_{i}b_{i}d_{i}-\sum _{1\leq i<j\leq n}(a_{i}d_{i}b_{j}c_{j}+a_{j}d_{j}b_{i}c_{i})-\sum _{i=1}^{n}a_{i}d_{i}b_{i}c_{i}\end{aligned}}}$$

where the second and fourth terms are the same and artificially added to complete the sums as follows:

$${\displaystyle =\sum _{i=1}^{n}\sum _{j=1}^{n}a_{i}c_{i}b_{j}d_{j}-\sum _{i=1}^{n}\sum _{j=1}^{n}a_{i}d_{i}b_{j}c_{j}.}$$

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

$${\displaystyle \det(AB)=\sum _{S\subset \{1,\ldots ,n\} \atop |S|=m}\det(A_{S})\det(B_{S}),}$$

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

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

Notes

References

• Aitken, Alexander Craig (1944), Determinants and Matrices, Oliver and Boyd
• Harville, David A. (2008), Matrix Algebra from a Statistician's Perspective, Springer