Lagrange's identity

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In mathematics, Lagrange's identity is an algebraic equation which applies to any two sets of numbers \{a_1, a_2,\dots, a_k\} and \{b_1, b_2,\dots, b_k\}. The identity is

\left( \sum_{k=1}^n a_k^2\right) \left(\sum_{k=1}^n b_k^2\right) - \left(\sum_{k=1}^n a_k b_k\right)^2 = \sum_{i=1}^{n-1} \sum_{j=i+1}^n (a_i b_j - a_j b_i)^2 \left(= {1 \over 2} \sum_{i=1}^n \sum_{j=1}^n (a_i b_j - a_j b_i)^2\right).

The identity is a special form of the Binet-Cauchy identity. Also, omitting the right side yields Cauchy's inequality.

[edit] Lagrange's identity and exterior algebra

In terms of the wedge product, Lagrange's identity can be written

(a \cdot a)(b \cdot b) - (a \cdot b)^2 = (a \wedge b) \cdot (a \wedge b).

Hence, it can be seen as a formula which gives the length of the wedge product of two vectors, which is the area of the paralleogram they define, in terms of the dot products of the two vectors, as

||a \wedge b|| = \sqrt{(||a||\ ||b||)^2 - ||a \cdot b||^2}.

[edit] Proof

The expansion of the first term on the left side is

\left( \sum_{k=1}^n a_k^2\right) \left(\sum_{k=1}^n b_k^2\right) = \sum_{i=1}^n \sum_{j=1}^n a_i^2 b_j^2 = \sum_{k=1}^n a_k^2 b_k^2 + \sum_{i=1}^{n-1} \sum_{j=i+1}^n a_i^2 b_j^2 + \sum_{j=1}^{n-1} \sum_{i=j+1}^n a_i^2 b_j^2 \qquad (1)

which means that the product of a column of as and a row of bs yields (a sum of elements of) a square of abs which can be broken up into a diagonal and a pair of triangles on either side of the diagonal.

The second term on the left side of Lagrange's identity can be expanded like so

\left(\sum_{k=1}^n a_k b_k\right)^2 = \sum_{k=1}^n a_k^2 b_k^2 + 2\sum_{i=1}^{n-1} \sum_{j=i+1}^n a_i b_i a_j b_j \qquad \qquad (2)

which means that a symmetric square can be broken up into its diagonal and a pair of equal triangles on either side of the diagonal.

To expand the summation on the right side of Lagrange's identity, first expand the square within the summation:

\sum_{i=1}^{n-1} \sum_{j=i+1}^n (a_i b_j - a_j b_i)^2 = \sum_{i=1}^{n-1} \sum_{j=i+1}^n (a_i^2 b_j^2 + a_j^2 b_i^2 - 2 a_i b_j a_j b_i).

Distribute the summation on the right side,

\sum_{i=1}^{n-1} \sum_{j=i+1}^n (a_i b_j - a_j b_i)^2 = \sum_{i=1}^{n-1} \sum_{j=i+1}^n a_i^2 b_j^2 + \sum_{i=1}^{n-1} \sum_{j=i+1}^n a_j^2 b_i^2 - 2 \sum_{i=1}^{n-1} \sum_{j=i+1}^n a_i b_j a_j b_i .

Now exchange the indices i and j of the second term on the right side, and permute the b factors of the third term, yielding

\sum_{i=1}^{n-1} \sum_{j=i+1}^n (a_i b_j - a_j b_i)^2 = \sum_{i=1}^{n-1} \sum_{j=i+1}^n a_i^2 b_j^2 + \sum_{j=1}^{n-1} \sum_{i=j+1}^n a_i^2 b_j^2 - 2 \sum_{i=1}^{n-1} \sum_{j=i+1}^n a_i b_i a_j b_j . \qquad (3)

Back to the left side of Lagrange's identity: it has two terms, given in expanded form by Equations (1) and (2). The first term on the right side of Equation (2) ends up cancelling out the first term on the right side of Equation (1), yielding

(1) - (2) = \sum_{i=1}^{n-1} \sum_{j=i+1}^n a_i^2 b_j^2 + \sum_{j=1}^{n-1} \sum_{i=j+1}^n a_i^2 b_j^2 - 2\sum_{i=1}^{n-1} \sum_{j=i+1}^n a_i b_i a_j b_j

which is the same as Equation (3), so Lagrange's identity is indeed an identity, q. e. d..

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