Binet–Cauchy identity

In algebra, the Binet–Cauchy identity, named after Jacques Philippe Marie Binet and Augustin-Louis Cauchy, states that
$\left(\sum_^n a_i c_i\right) \left(\sum_^n b_j d_j\right) = \left(\sum_^n a_i d_i\right) \left(\sum_^n b_j c_j\right) + \sum_ (a_i b_j - a_j b_i ) (c_i d_j - c_j d_i )$
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 $(a \wedge b) \cdot (c \wedge d) = (a \cdot c)(b \cdot d) - (a \cdot d)(b \cdot c)$ form of the Binet–Cauchy identity can be written as
$\frac\left(\varepsilon^ ~ a_ ~ b_ \right)\left( \varepsilon_ ~ c^ ~ d^\right) = \delta^_ ~ a_ ~ b_ ~ c^ ~ d^\,.$

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
$A = \begina_1&\dots&a_n\\b_1&\dots& b_n\end,\quad B = \beginc_1&d_1\\\vdots&\vdots\\c_n&d_n\end.$

Notes

References

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{{DEFAULTSORT:Binet-Cauchy Identity
Algebraic identities
Multilinear algebra
Articles containing proofs