Binet–Cauchy identity

In algebra, the Binet–Cauchy identity, named after Jacques Philippe Marie Binet and Augustin-Louis Cauchy, states that $$ \left(\sum_{i=1}^n a_i c_i\right) \left(\sum_{j=1}^n b_j d_j\right) = \left(\sum_{i=1}^n a_i d_i\right) \left(\sum_{j=1}^n b_j c_j\right) + \sum_{1\le i < j \le n} (a_i b_j - a_j b_i ) (c_i d_j - c_j d_i ) $$ 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 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 $n = 3$, 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 $$(a \wedge b) \cdot (c \wedge d) = (a \cdot c)(b \cdot d) - (a \cdot d)(b \cdot c)\,,$$ 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 $n = 3$ case.

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

When both $a$ and $b$ are unit vectors, we obtain the usual relation $$\sin^2 \phi = 1 - \cos^2 \phi$$ 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 $$\frac{1}{k!}\varepsilon^{\lambda_1\cdots\lambda_k\mu_{k+1}\cdots\mu_{n}} \varepsilon_{\lambda_1\cdots\lambda_k\nu_{k+1}\cdots\nu_{n}} = \delta^{\mu_{k+1}\cdots\mu_{n}}_{\nu_{k+1}\cdots\nu_{n}}\,.$$

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 $$\frac{1}{(n-2)!}\left(\varepsilon^{\mu_1\cdots\mu_{n-2}\alpha\beta} ~ a_{\alpha} ~ b_{\beta} \right)\left( \varepsilon_{\mu_1\cdots\mu_{n-2}\gamma\delta} ~ c^{\gamma} ~ d^{\delta}\right) = \delta^{\alpha\beta}_{\gamma\delta} ~ a_{\alpha} ~ b_{\beta} ~ c^{\gamma} ~ d^{\delta}\,.$$

Proof
Expanding the last term, $$ \begin{align} &\sum_{1\le i < j \le n} (a_i b_j - a_j b_i ) (c_i d_j - c_j d_i ) \\ ={}&{} \sum_{1\le i < j \le n} (a_i c_i b_j d_j + a_j c_j b_i d_i) + \sum_{i=1}^n a_i c_i b_i d_i - \sum_{1\le i < j \le n} (a_i d_i b_j c_j + a_j d_j b_i c_i) - \sum_{i=1}^n a_i d_i b_i c_i \end{align} $$ 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 AS for the m×m matrix whose columns are those columns of A that have indices from S. Similarly, we write BS 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 $$\det(AB) = \sum_{ S\subset\{1,\ldots,n\} \atop |S| = m} \det(A_S)\det(B_S),$$ 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 $$ A = \begin{pmatrix}a_1&\dots&a_n\\b_1&\dots& b_n\end{pmatrix},\quad B = \begin{pmatrix}c_1&d_1\\\vdots&\vdots\\c_n&d_n\end{pmatrix}. $$