Binet–Cauchy identity

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

$\biggl(\sum_{i=1}^n a_i c_i\biggr) \biggl(\sum_{j=1}^n b_j d_j\biggr) = \biggl(\sum_{i=1}^n a_i d_i\biggr) \biggl(\sum_{j=1}^n b_j c_j\biggr) + \sum_{1\le i < j \le n} (a_i b_j - a_j b_i ) (c_i d_j - c_j d_i )$

for every choice of real or complex numbers (or more generally, elements of a commutative ring). Setting a_i = c_i and b_i = d_i, it gives the Lagrange's identity, which is a stronger version of the Cauchy-Schwarz inequality for the Euclidean space $\scriptstyle\mathbb{R}^n$ .

[edit] 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

$(a \wedge b) \cdot (c \wedge d) = (a \cdot c)(b \cdot d) - (a \cdot d)(b \cdot c).\,$

[edit] Proof

Expanding the last term,

$\sum_{1\le i < j \le n} (a_i b_j - a_j b_i ) (c_i d_j - c_j d_i )$

$= \sum_{1\le i < j \le n} (a_i c_i b_j d_j + a_j c_j b_i d_i) +\sum_{i=1}^n a_i c_i b_i d_i - \sum_{1\le i < j \le n} (a_i d_i b_j c_j + a_j d_j b_i c_i) - \sum_{i=1}^n a_i d_i b_i c_i$