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

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

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 $\mathbb{R}^n$ .

[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,

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

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