Lagrange’s identity

Let R be a commutative ring, and let x1,,xn,y1,,yn be arbitrary elements in R. Then


Since R is commutativePlanetmathPlanetmathPlanetmath, we can apply the binomial formula.We start out with

(i=1nxiyi)2=i=1n(xi2yi2)+1i<jn2xiyjxjyi (1)

Using the binomial formula, we see that


So we get

(i=1nxiyi)2+1i<jnn(xiyj-xjyi)2 = i=1n(xi2yi2)+1i<jnn(xi2yj2+xj2yi2) (2)
= (i=1nxi2)(i=1nyi2) (3)

Note that changing the roles of i and j in xiyj-xjyi, we get


but the negative sign will disappear when we square. So we can rewrite the last equation to

(i=1nxiyi)2+1i<jn(xiyj-xjyi)2=(i=1nxi2)(i=1nyi2). (4)

This is equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmath to the stated identityPlanetmathPlanetmathPlanetmathPlanetmath. ∎

Title Lagrange’s identity
Canonical name LagrangesIdentity
Date of creation 2013-03-22 13:18:01
Last modified on 2013-03-22 13:18:01
Owner mathcam (2727)
Last modified by mathcam (2727)
Numerical id 21
Author mathcam (2727)
Entry type Theorem
Classification msc 13A99