PlanetMath: Lagrange's identity

Let

be a commutative ring, and let $x_1, \ldots, x_n, y_1, \ldots, y_n$ be arbitrary elements in

. Then

Proof. Since

is commutative, we can apply the binomial formula.We start out with

$\displaystyle \left(\sum_{i=1}^n x_iy_i\right)^2 =\sum_{i=1}^n (x_i^2y_i^2) +\sum_{1 \leq i< j\leq n} 2 x_iy_jx_jy_i$

(1)

Using the binomial formula, we see that

$\displaystyle (x_iy_j -x_jy_i)^2 =x_i^2y_j^2 -2x_ix_jy_iy_j +x_j^2y_i^2.$

So we get

$\displaystyle \left(\sum\limits_{i=1}^n x_iy_i\right)^2 +\sum\limits_{1 \leq i< j\leq n}^n(x_iy_j -x_jy_i)^2$	$\displaystyle =$	$\displaystyle \sum\limits_{i=1}^n \left(x_i^2y_i^2\right) +\sum\limits_{1 \leq i< j\leq n}^n \left(x_i^2y_j^2 +x_j^2y_i^2\right)$	(2)
	$\displaystyle =$	$\displaystyle \left(\sum\limits_{i=1}^n x_i^2\right)\left(\sum\limits_{i=1}^n y_i^2\right)$	(3)

Note that changing the roles of

and

, we get

$\displaystyle x_jy_i -x_iy_j =-(x_iy_j -x_jy_i),$

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

$\displaystyle \left(\sum\limits_{i=1}^n x_iy_i\right)^2 +\sum\limits_{1 \le i <... ...y_j -x_jy_i)^2 =\left(\sum_{i=1}^n x_i^2\right)\left(\sum_{i=1}^n y_i^2\right).$

(4)

This is equivalent to the stated identity. $\qedsymbol$