View source for Lagrange's identity

{{short description|On products on sums of squares}}
{{other uses|Lagrange's identity (disambiguation)}}

In the [[algebra]], '''Lagrange's identity''', named after [[Joseph Louis Lagrange]], is:<ref name=Weisstein>{{cite book |title=CRC concise encyclopedia of mathematics |author=Eric W. Weisstein |year= 2003|url=https://books.google.com/books?id=8LmCzWQYh_UC&pg=PA228 |isbn=1-58488-347-2 |edition=2nd |publisher=CRC Press }}</ref><ref name= Greene>{{cite book |title=Function theory of one complex variable |author1=Robert E Greene|author1-link= Robert Everist Greene |author2=Steven G Krantz |page=22 |chapter=Exercise 16   |isbn=0-8218-3962-4 |year=2006 |edition=3rd |publisher=American Mathematical Society}}</ref>
<math display="block">\begin{align}
\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 \left(a_i b_j - a_j b_i\right)^2 \\
& \left(= \frac{1}{2} \sum_{i=1}^n \sum_{j=1,j\neq i}^n (a_i b_j - a_j b_i)^2\right),
\end{align}</math>
which applies to any two sets {''a''<sub>1</sub>, ''a''<sub>2</sub>, ..., ''a<sub>n</sub>''} and {''b''<sub>1</sub>, ''b''<sub>2</sub>, ..., ''b<sub>n</sub>''} of [[real number|real]] or [[complex number]]s (or more generally, elements of a [[commutative ring]]). This identity is a generalisation of the [[Brahmagupta–Fibonacci identity]] and a special form of the [[Binet–Cauchy identity]].

In a more compact vector notation, Lagrange's identity is expressed as:<ref name=Boichenko>{{cite book |title=Dimension theory for ordinary differential equations |author1=Vladimir A. Boichenko |author2=Gennadiĭ Alekseevich Leonov |author3=Volker Reitmann |url=https://books.google.com/books?id=9bN1-b_dSYsC&pg=PA26 |page=26 |isbn=3-519-00437-2 |year=2005 |publisher=Vieweg+Teubner Verlag}}</ref>
<math display="block">\left\| \mathbf a \right\|^2 \left\| \mathbf b \right\|^2 - (\mathbf{a} \cdot \mathbf{b} )^2 = \sum_{1 \le i < j \le n} \left(a_i b_j - a_j b_i \right)^2 \, , </math>
where '''a''' and '''b''' are ''n''-dimensional vectors with components that are real numbers. The extension to complex numbers requires the interpretation of the [[dot product]] as an [[inner product]] or Hermitian dot product. Explicitly, for complex numbers, Lagrange's identity can be written in the form:<ref name=Steele>{{cite book |title=The Cauchy-Schwarz master class: an introduction to the art of mathematical inequalities |author=J. Michael Steele |pages=68–69 |chapter=Exercise 4.4: Lagrange's identity for complex numbers |chapter-url=https://books.google.com/books?id=bvgBdZKEYAEC&pg=PA68  |year=2004 |isbn=0-521-54677-X |publisher=Cambridge University Press}}</ref>
<math display="block">\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 \left|a_i \overline{b}_j - a_j \overline{b}_i\right|^2</math>
involving the [[Absolute value#Complex numbers|absolute value]].<ref>{{Cite book | last1=Greene | first1=Robert E. | last2=Krantz | first2=Steven G. | title=Function Theory of One Complex Variable | publisher=[[American Mathematical Society]] | location=Providence, R.I. | isbn=978-0-8218-2905-9 | year=2002 | page=22, Exercise 16 }}</ref><ref>{{Cite book | last1=Palka | first1=Bruce P. | title=An Introduction to Complex Function Theory | url=https://archive.org/details/introductiontoco00palk_656 | url-access=limited | publisher=[[Springer-Verlag]] | location=Berlin, New York | isbn=978-0-387-97427-9 | year=1991 | page=[https://archive.org/details/introductiontoco00palk_656/page/n45 27], Exercise 4.22 }}.</ref>

Since the right-hand side of the identity is clearly non-negative, it implies [[Cauchy–Schwarz inequality|Cauchy's inequality]] in the [[Dimension (vector space)|finite-dimensional]] [[real coordinate space]] '''R'''<sup>''n''</sup> and its complex counterpart '''C'''<sup>''n''</sup>.

Geometrically, the identity asserts that the square of the volume of the parallelepiped spanned by a set of vectors is the [[Gramian|Gram determinant]] of the vectors.

==Lagrange's identity and exterior algebra==

In terms of the [[wedge product]], Lagrange's identity can be written
<math display="block">(a \cdot a)(b \cdot b) - (a \cdot b)^2 = (a \wedge b) \cdot (a \wedge b).</math>

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 parallelogram they define, in terms of the dot products of the two vectors, as
<math display="block">\|a \wedge b\| = \sqrt{(a \cdot a)(b \cdot b) - (a \cdot b)^2} = \sqrt{\|a\|^2\|b\|^2 - (a \cdot b)^2}.</math>

==Lagrange's identity and vector calculus==

In three dimensions, Lagrange's identity asserts that if '''a''' and '''b''' are vectors in '''R'''<sup>3</sup> with lengths |'''a'''| and |'''b'''|, then Lagrange's identity can be written in terms of the [[cross product]] and [[dot product]]:<ref name=Anton>{{cite book |title=Elementary Linear Algebra: Applications Version |author1=Howard Anton |author2=Chris Rorres |chapter-url=https://books.google.com/books?id=1PJ-WHepeBsC&dq=%22cross+product%22+%22Lagrange%27s+identity%22&pg=PA162 |page=162 |chapter=Relationships between dot and cross products |isbn=978-0-470-43205-1 |year=2010 |publisher=John Wiley and Sons |edition=10th}}</ref><ref name=Lounesto1>{{cite book |title=Clifford algebras and spinors |author=Pertti Lounesto |url=https://books.google.com/books?id=kOsybQWDK4oC&dq=%22which+in+coordinate+form+means+Lagrange%27s+identity%22&pg=PA94 |isbn=0-521-00551-5 |year=2001 |page=94 |edition=2nd |publisher=Cambridge University Press}}</ref>
<math display="block"> |\mathbf{a}|^2 |\mathbf{b}|^2 - (\mathbf {a} \cdot \mathbf{b})^2 = |\mathbf a \times \mathbf b|^2</math>

Using the definition of angle based upon the [[Dot product#Geometric interpretation|dot product]] (see also [[Cauchy–Schwarz inequality#Applications|Cauchy–Schwarz inequality]]), the left-hand side is
<math display="block">\left|\mathbf{a}\right|^2 \left|\mathbf{b}\right|^2 \left( 1- \cos^2\theta\right) = \left|\mathbf{a}\right|^2 \left|\mathbf{b}\right|^2\sin^2\theta</math>
where {{mvar|θ}} is the angle formed by the vectors '''a''' and '''b'''.  The area of a parallelogram with sides {{math|{{abs|'''a'''}}}} and {{math|{{abs|'''b'''}}}} and angle {{mvar|θ}} is known in elementary geometry to be
<math display="block">\left|\mathbf{a}\right| \left|\mathbf{b}\right| \left|\sin\theta\right|,</math>
so the left-hand side of Lagrange's identity is the squared area of the parallelogram.  The cross product appearing on the right-hand side is defined by
<math display="block">\mathbf{a}\times\mathbf{b} = \left(a_2 b_3- a_3 b_2\right)\mathbf{i} + \left(a_3 b_1 - a_1 b_3\right)\mathbf{j} + \left(a_1 b_2 - a_2 b_1\right)\mathbf{k}</math>
which is a vector whose components are equal in magnitude to the areas of the projections of the parallelogram onto the ''yz'', ''zx'', and ''xy'' planes, respectively.

===Seven dimensions===
{{main|Seven-dimensional cross product}} 
For '''a''' and '''b''' as vectors in '''R'''<sup>7</sup>, Lagrange's identity takes on the same form as in the case of '''R'''<sup>3</sup><ref name=Lounesto>{{cite book |title=Clifford algebras and spinors |author=Door Pertti Lounesto |isbn=0-521-00551-5 |year=2001 |edition = 2nd |url=https://books.google.com/books?id=kOsybQWDK4oC&q=Pythagorean |publisher=Cambridge University Press}} See particularly [https://books.google.com/books?id=kOsybQWDK4oC&pg=PA96 § 7.4 Cross products in '''R'''<sup>7</sup>], p. 96.</ref>
<math display="block">|\mathbf{a}|^2 |\mathbf{b}|^2 -|\mathbf{a} \cdot \mathbf{b}|^2 = |\mathbf{a} \times \mathbf{b}|^2 \ ,</math>

However, the cross product in 7 dimensions does not share all the properties of the cross product in 3 dimensions. For example, the direction of '''a''' × '''b''' in 7-dimensions may be the same as '''c''' × '''d''' even though '''c''' and '''d''' are linearly independent of '''a''' and '''b'''. Also the [[seven-dimensional cross product]] is not compatible with the [[Jacobi identity]].<ref name=Lounesto/>

===Quaternions===
A [[quaternion]] ''p'' is defined as the sum of a scalar ''t'' and a vector '''v''':
<math display="block">p = t + \mathbf v = t + x \ \mathbf  i +y \ \mathbf j + z\  \mathbf k.</math>

The product of two quaternions {{math|1=''p'' = ''t'' + '''v'''}} and {{math|1=''q'' = ''s'' + '''w'''}} is defined by
<math display="block">pq = (st - \mathbf{v}\cdot\mathbf{w}) + s \ \mathbf{v} + t \ \mathbf{w}  + \mathbf{v}\times\mathbf{w}.</math>

The quaternionic conjugate of ''q'' is defined by
<math display="block">\overline{q} = t - \mathbf{v},</math>
and the norm squared is
<math display="block">|q|^2 = q\overline{q} = t^2 \ + \ x ^2   + \ y^2 \ +\  z^2.</math>

The multiplicativity of the norm in the quaternion algebra provides, for quaternions ''p'' and ''q'':<ref name=Kuipers>{{cite book |title=Quaternions and rotation sequences: a primer with applications to orbits |author=Jack B. Kuipers |chapter-url=https://books.google.com/books?id=_2sS4mC0p-EC&pg=PA111 |page=111 |chapter=§5.6 The norm |isbn=0-691-10298-8 |year=2002 |publisher=Princeton University Press}}</ref>
<math display="block">\left|pq\right| = \left|p\right| \left|q\right|.</math>

The quaternions ''p'' and ''q'' are called imaginary if their scalar part is zero; equivalently, if
<math display="block">p = \mathbf{v},\quad q = \mathbf{w}.</math>

Lagrange's identity is just the multiplicativity of the norm of imaginary quaternions,
<math display="block">|\mathbf{v}\mathbf{w}|^2 = |\mathbf{v}|^2|\mathbf{w}|^2,</math>
since, by definition,
<math display="block">|\mathbf{v}\mathbf{w}|^2 = (\mathbf{v}\cdot\mathbf{w})^2 + |\mathbf{v}\times\mathbf{w}|^2.</math>

== Proof of algebraic form ==

The vector form follows from the Binet-Cauchy identity by setting ''c<sub>i</sub>''&nbsp;=&nbsp;''a<sub>i</sub>'' and ''d<sub>i</sub>''&nbsp;=&nbsp;''b<sub>i</sub>''. The second version follows by letting ''c<sub>i</sub>'' and ''d<sub>i</sub>'' denote the [[complex conjugate]]s of ''a<sub>i</sub>'' and ''b<sub>i</sub>'', respectively,

Here is also a direct proof.<ref name=Jones>See, for example, [https://docs.google.com/viewer?a=v&q=cache:rDnOA-ZKljkJ:www.owlnet.rice.edu/~fjones/chap7.pdf+lagrange%27s+identity+in+the+seven+dimensional+cross+product&hl=en&gl=ph&sig=AHIEtbQQtdVGhgbYhz78SQQb2biLxRi4kA Frank Jones, Rice University], page 4 in Chapter 7 of a [http://www.owlnet.rice.edu/~fjones/ book still to be published].</ref> The expansion of the first term on the left side is:
{{NumBlk||<math display="block"> \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 \ ,</math>|{{EquationRef|1}}}}
which means that the product of a column of ''a''<small>s</small> and a row of ''b''<small>s</small> yields (a sum of elements of) a square of ''ab''<small>s</small>, 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 as:
{{NumBlk||<math display="block"> \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 \, ,</math>|{{EquationRef|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:
<math display="block"> \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 \left(a_i^2 b_j^2 + a_j^2 b_i^2 - 2 a_i b_j a_j b_i\right). </math>

Distribute the summation on the right side,
<math display="block"> \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 .</math>

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:
{{NumBlk||<math display="block"> \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 \ .</math>|{{EquationRef|3}}}}

Back to the left side of Lagrange's identity: it has two terms, given in expanded form by Equations ({{EquationNote|1}}) and ({{EquationNote|2}}).  The first term on the right side of Equation ({{EquationNote|2}}) ends up canceling out the first term on the right side of Equation ({{EquationNote|1}}), yielding
{{NumBlk||<math display="block">\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 </math>
|3='''('''<nowiki/>{{EquationNote|1}}<nowiki/>''')''' − '''('''<nowiki/>{{EquationNote|2}}<nowiki/>''')'''|RawN=yes}}
which is the same as Equation ({{EquationNote|3}}), so Lagrange's identity is indeed an identity, [[Q.E.D.]]

== Proof of Lagrange's identity for complex numbers ==
Normed division algebras require that the norm of the product is equal to the product of the norms. Lagrange's identity exhibits this equality.
The product identity used as a starting point here, is a consequence of the norm of the product equality with the product of the norm for scator algebras. This proposal, originally presented in the context of a deformed Lorentz metric, is based on a transformation stemming from the product operation and magnitude definition in hyperbolic scator algebra.<ref>M. Fernández-Guasti, ''Alternative realization for the composition of relativistic velocities'', Optics and Photonics 2011, vol. 8121 of The nature of light: What are photons? IV, pp. 812108–1–11. SPIE, 2011.</ref>
Lagrange's identity can be proved in a variety of ways.<ref name="Steele"/>

Let <math>a_{i},b_{i}\in\mathbb{C}</math> be complex numbers and the overbar represents complex conjugate.

The product identity <math display="block">\prod_{i=1}^{n} \left(1 - a_{i} \bar{a}_{i} - b_{i} \bar{b}_{i} + a_{i} \bar{a}_{i} b_{i} \bar{b}_{i}\right) = \prod_{i=1}^{n} \left(1 - a_{i} \bar{a}_{i}\right) \prod_{i=1}^{n} \left(1 - b_{i} \bar{b}_{i}\right)</math> reduces to the complex Lagrange's identity when fourth order terms, in a series expansion, are considered.

In order to prove it, expand the product on the LHS of the product identity in terms of
series up to fourth order. To this end, recall that products of the form <math>\left(1 +x_{i}\right)</math> can be expanded in terms of sums as
<math display="block">\prod_{i=1}^{n}\left(1+x_{i}\right) = 1+\sum_{i=1}^{n}x_{i}+\sum_{i<j}^{n}x_{i}x_{j}+\mathcal{O}^{3+}(x),</math>
where <math>\mathcal{O}^{3+}(x)</math> means terms with order three or higher in <math>x</math>.

<math display="block">
\prod_{i=1}^{n}\left(1 - a_{i} \bar{a}_{i} - b_{i} \bar{b}_{i} + a_{i} \bar{a}_{i} b_{i} \bar{b}_{i}\right) = 1-\sum_{i=1}^{n} \left(a_{i}\bar{a}_{i} + b_{i} \bar{b}_{i}\right) + \sum_{i=1}^{n} a_{i} \bar{a}_{i} b_{i} \bar{b}_{i}
+ \sum_{i<j}^{n} \left(a_{i} \bar{a}_{i} a_{j} \bar{a}_{j} + b_{i} \bar{b}_{i} b_{j} \bar{b}_{j}\right)+\sum_{i<j}^{n}\left(a_{i} \bar{a}_{i} b_{j} \bar{b}_{j} + a_{j} \bar{a}_{j} b_{i} \bar{b}_{i}\right) + \mathcal{O}^{5+}.
</math>

The two factors on the RHS are also written in terms of series
<math display="block">
\prod_{i=1}^{n}\left(1-a_{i}\bar{a}_{i}\right)\prod_{i=1}^{n}\left(1-b_{i}\bar{b}_{i}\right)=\left(1-\sum_{i=1}^{n}a_{i}\bar{a}_{i}+\sum_{i<j}^{n}a_{i}\bar{a}_{i}a_{j}\bar{a}_{j}+\mathcal{O}^{5+}\right)
\left(1-\sum_{i=1}^{n}b_{i}\bar{b}_{i}+\sum_{i<j}^{n}b_{i}\bar{b}_{i}b_{j}\bar{b}_{j}+\mathcal{O}^{5+}\right).
</math>

The product of this expression up to fourth order is 
<math display="block">
\prod_{i=1}^{n}\left(1-a_{i}\bar{a}_{i}\right)\prod_{i=1}^{n}\left(1-b_{i}\bar{b}_{i}\right)=1-\sum_{i=1}^{n}\left(a_{i}\bar{a}_{i}+b_{i}\bar{b}_{i}\right)
+\left(\sum_{i=1}^{n}a_{i}\bar{a}_{i}\right)\left(\sum_{i=1}^{n}b_{i}\bar{b}_{i}\right)+\sum_{i<j}^{n}\left(a_{i}\bar{a}_{i}a_{j}\bar{a}_{j}+b_{i}\bar{b}_{i}b_{j}\bar{b}_{j}\right)+\mathcal{O}^{5+}.
</math>
Substitution of these two results in the product identity give 
<math display="block">
\sum_{i=1}^{n}a_{i}\bar{a}_{i}b_{i}\bar{b}_{i}+\sum_{i<j}^{n}\left(a_{i}\bar{a}_{i}b_{j}\bar{b}_{j}+a_{j}\bar{a}_{j}b_{i}\bar{b}_{i}\right)=\left(\sum_{i=1}^{n}a_{i}\bar{a}_{i}\right)\left(\sum_{i=1}^{n}b_{i}\bar{b}_{i}\right).
</math>

The product of two conjugates series can be expressed as series involving the product of conjugate terms. The conjugate series product is <math display="block">\left(\sum_{i=1}^{n} x_{i}\right)\left(\sum_{i=1}^{n}\bar{x}_{i}\right) = \sum_{i=1}^{n} x_{i}\bar{x}_{i}+\sum_{i<j}^{n}\left(x_{i}\bar{x}_{j}+\bar{x}_{i}x_{j}\right),</math> thus
<math display="block">
\left(\sum_{i=1}^{n}a_{i}b_{i}\right)\left(\sum_{i=1}^{n}\overline{a_{i}b_{i}}\right)-\sum_{i<j}^{n}\left(a_{i}b_{i}\bar{a}_{j}\bar{b}_{j}+\bar{a}_{i}\bar{b}_{i}a_{j}b_{j}\right)+\sum_{i<j}^{n}\left(a_{i}\bar{a}_{i}b_{j}\bar{b}_{j}+a_{j}\bar{a}_{j}b_{i}\bar{b}_{i}\right)
=\left(\sum_{i=1}^{n}a_{i}\bar{a}_{i}\right)\left(\sum_{i=1}^{n}b_{i}\bar{b}_{i}\right).
</math>

The terms of the last two series on the LHS are grouped as
<math display="block">a_{i}\bar{a}_{i}b_{j}\bar{b}_{j}+a_{j}\bar{a}_{j}b_{i}\bar{b}_{i}-a_{i}b_{i}\bar{a}_{j}\bar{b}_{j}-\bar{a}_{i}\bar{b}_{i}a_{j}b_{j} = \left(a_{i}\bar{b}_{j}-a_{j}\bar{b}_{i}\right)\left(\bar{a}_{i}b_{j}-\bar{a}_{j}b_{i}\right),</math> in order to obtain the complex Lagrange's identity:
<math display="block">
\left(\sum_{i=1}^{n}a_{i}b_{i}\right)\left(\sum_{i=1}^{n}\overline{a_{i}b_{i}}\right)+\sum_{i<j}^{n}\left(a_{i}\bar{b}_{j}-a_{j}\bar{b}_{i}\right)\left(\overline{a_{i}\bar{b}_{j}-a_{j}\bar{b}_{i}}\right)=\left(\sum_{i=1}^{n}a_{i}\bar{a}_{i}\right)\left(\sum_{i=1}^{n}b_{i}\bar{b}_{i}\right).
</math>

In terms of the moduli,
<math display="block">
\left|\sum_{i=1}^{n}a_{i}b_{i}\right|^{2}+\sum_{i<j}^{n}\left|a_{i}\bar{b}_{j}-a_{j}\bar{b}_{i}\right|^{2}=\left(\sum_{i=1}^{n}\left|a_{i}\right|^{2}\right)\left(\sum_{i=1}^{n}\left|b_{i}\right|^{2}\right).
</math>

Lagrange's identity for complex numbers has been obtained from a straightforward
product identity. A derivation for the reals is obviously even more succinct. Since the Cauchy–Schwarz inequality is a particular case of Lagrange's identity,<ref name="Steele"/> this
proof is yet another way to obtain the CS inequality. Higher order terms in the series produce novel identities.

== See also ==
*[[Brahmagupta–Fibonacci identity]]
*[[Lagrange's identity (boundary value problem)]]
*[[Binet–Cauchy identity]]

== References ==

<references/>

== External links ==
*{{MathWorld|LagrangesIdentity|Lagrange's Identity}}

{{Joseph-Louis Lagrange}}

[[Category:Algebraic identities]]
[[Category:Multilinear algebra]]
[[Category:Articles containing proofs]]