Binet–Cauchy identity explained

In algebra, the Binet–Cauchy identity, named after Jacques Philippe Marie Binet and Augustin-Louis Cauchy, states that^[1] $\left(\sum_^n a_i c_i\right)\left(\sum_^n b_j d_j\right) = \left(\sum_^n a_i d_i\right)\left(\sum_^n b_j c_j\right) + \sum_ (a_i b_j - a_j b_i) (c_i d_j - c_j d_i)$ for every choice of real or complex numbers (or more generally, elements of a commutative ring).Setting and, 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, the first and second terms on the right hand side become the squared magnitudes of dot and cross products respectively; in 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,,, and 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 case.

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

When both and are unit vectors, we obtain the usual relation $\sin^2 \phi = 1 - \cos^2 \phi$ where 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 $\frac\varepsilon^ \varepsilon_ = \delta^_\,.$

The

(a\wedgeb) ⋅ (c\wedged)=(a ⋅ c)(b ⋅ d)-(a ⋅ d)(b ⋅ c)

form of the Binet–Cauchy identity can be written as

\frac\left(\varepsilon^ ~ a_ ~ b_ \right)\left(\varepsilon_ ~ c^ ~ d^\right) = \delta^_ ~ a_ ~ b_ ~ c^ ~ d^\,.

Proof

Expanding the last term, $\begin&\sum_ (a_i b_j - a_j b_i) (c_i d_j - c_j d_i) \\=&\sum_ (a_i c_i b_j d_j + a_j c_j b_i d_i)+ \sum_^n a_i c_i b_i d_i- \sum_ (a_i d_i b_j c_j + a_j d_j b_i c_i)- \sum_^n a_i d_i b_i c_i\end$ where the second and fourth terms are the same and artificially added to complete the sums as follows: $=\sum_^n \sum_^n a_i c_i b_j d_j-\sum_^n \sum_^n a_i d_i b_j c_j.$

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 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 $\det(AB) = \sum_ \det(A_S)\det(B_S),$ where the sum extends over all possible subsets S of with m elements.

We get the original identity as special case by setting $A = \begina_1&\dots&a_n\\b_1&\dots& b_n\end,\quadB = \beginc_1&d_1\\\vdots&\vdots\\c_n&d_n\end.$

Notes

Book: CRC concise encyclopedia of mathematics . Eric W. Weisstein . 228 . https://books.google.com/books?id=8LmCzWQYh_UC&pg=PA228 . Binet-Cauchy identity . 1-58488-347-2 . 2003 . 2nd . CRC Press.