Lagrange's identity

In algebra, Lagrange's identity is the identity

:

which applies to any two sets {"a"₁, "a"₂, . . ., "a_n"} and {"b"₁, "b"₂, . . ., "b_n"} of real or complex numbers (or more generally, elements of a commutative ring). This identity is a special form of the Binet–Cauchy identity. For complex numbers it can also be written in the form

:

involving the absolute value.

Since the right-hand side of the identity is clearly non-negative, it implies Cauchy's inequality in the finite-dimensional real coordinate space $mathbb\{R\}^n$ and its complex counterpart $mathbb\{C\}^n$.

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

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

Lagrange's identity and vector calculus

If "a" and "b" are vectors in $mathbb\{R\}^3$, Lagrange's identity can be also written in terms of the cross product and dot product:

:$|a\; imes\; b|^2\; +\; (a\; cdot\; b)^2\; =\; |a|^2\; |b|^2\; =\; (a\; cdot\; a)(b\; cdot\; b).$

This is a special case of the multiplicativity of the norm in the quaternion algebra:

:$|vw|\; =\; |v|\; |w|.\; ,$

Or more generally,

:$(v\; imes\; w)\; cdot\; (a\; imes\; b)\; =\; (v\; cdot\; a)(w\; cdot\; b)\; -\; (v\; cdot\; b)(w\; cdot\; a).$

Lagrange's identity and calculus

In terms of the Sturm-Liouville theory, Lagrange's identity can be writtenwhere $p=P(x)$, $q=Q(x)$, $u=U(x)$ and $v=V(x)$ are functions of $x$. $u$ and $v$ having continuous second derivatives on the interval $[0,1]$. $L$ is Sturm-Liouville differential operators defined by

Proof

Algebraic form

The first version follows from the Binet-Cauchy identity by setting "c_i" = "a_i" and "d_i" = "b_i". The second version follows by letting "c_i" and "d_i" denote the complex conjugates of "a_i" and "b_i", respectively,

Here is also a direct proof of the first version. The expansion of the first term on the left side isNumBlk|1=:|2=$left(\; sum\_\{k=1\}^n\; a\_k^2\; ight)\; left(sum\_\{k=1\}^n\; b\_k^2\; ight)\; =\; sum\_\{i=1\}^n\; sum\_\{j=1\}^n\; a\_i^2\; b\_j^2\; =\; sum\_\{k=1\}^n\; a\_k^2\; b\_k^2\; +\; sum\_\{i=1\}^\{n-1\}\; sum\_\{j=i+1\}^n\; a\_i^2\; b\_j^2\; +\; sum\_\{j=1\}^\{n-1\}\; sum\_\{i=j+1\}^n\; a\_i^2\; b\_j^2$|3=$(3)$|RawN=.which means that the product of a column of "a"s and a row of "b"s yields (a sum of elements of) a square of "ab"s 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 soNumBlk|1=:|2=$left(sum\_\{k=1\}^n\; a\_k\; b\_k\; ight)^2\; =\; sum\_\{k=1\}^n\; a\_k^2\; b\_k^2\; +\; 2sum\_\{i=1\}^\{n-1\}\; sum\_\{j=i+1\}^n\; a\_i\; b\_i\; a\_j\; b\_j$|3=$(4)$|RawN=.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

Back to the left side of Lagrange's identity: it has two terms, given in expanded form by Equations (3) and (4). The first term on the right side of Equation (4) ends up cancelling out the first term on the right side of Equation (3), yielding:$(3)\; -\; (4)\; =\; sum\_\{i=1\}^\{n-1\}\; sum\_\{j=i+1\}^n\; a\_i^2\; b\_j^2\; +\; sum\_\{j=1\}^\{n-1\}\; sum\_\{i=j+1\}^n\; a\_i^2\; b\_j^2\; -\; 2sum\_\{i=1\}^\{n-1\}\; sum\_\{j=i+1\}^n\; a\_i\; b\_i\; a\_j\; b\_j$which is the same as Equation (5), so Lagrange's identity is indeed an identity, "q. e. d.".

= Calculus form =

Replace $f(x)=pu\text{'}$, $g(x)=v$, $a=0$ and $b=1$ into the rule integration by partswe haveReplace $f(x)=u$, $g(x)=pv\text{'}$, $a=0$ and $b=1$ into the rule (6) again, we have:Replace (8) into (7), we get:From the definition (2), we can getReplace (9) into (10), we have:Rearrange terms of (11) then (1) is obtained. q.e.d.

See also

Brahmagupta-Fibonacci_identity

References