Lagrange’s identity

Let $R$ be a commutative ring, and let $x_{1},\ldots,x_{n},y_{1},\ldots,y_{n}$ be arbitrary elements in $R$ . Then

$\left(\sum_{k=1}^{n}x_{k}y_{k}\right)^{2}=\left(\sum_{k=1}^{n}x_{k}^{2}\right)% \left(\sum_{k=1}^{n}y_{k}^{2}\right)-\sum_{1\leq k<i\leq n}(x_{k}y_{i}-x_{i}y_% {k})^{2}\mbox{.}$

Proof.

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

$\left(\sum_{i=1}^{n}x_{i}y_{i}\right)^{2}=\sum_{i=1}^{n}(x_{i}^{2}y_{i}^{2})+% \sum_{1\leq i<j\leq n}2x_{i}y_{j}x_{j}y_{i}$

(1)

Using the binomial formula, we see that

$(x_{i}y_{j}-x_{j}y_{i})^{2}=x_{i}^{2}y_{j}^{2}-2x_{i}x_{j}y_{i}y_{j}+x_{j}^{2}% y_{i}^{2}.$

So we get

	$\displaystyle\left(\sum\limits_{i=1}^{n}x_{i}y_{i}\right)^{2}+\sum\limits_{1% \leq i<j\leq n}^{n}(x_{i}y_{j}-x_{j}y_{i})^{2}$	$\displaystyle=$	$\displaystyle\sum\limits_{i=1}^{n}\left(x_{i}^{2}y_{i}^{2}\right)+\sum\limits_% {1\leq i<j\leq n}^{n}\left(x_{i}^{2}y_{j}^{2}+x_{j}^{2}y_{i}^{2}\right)$		(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 $i$ and $j$ in $x_{i}y_{j}-x_{j}y_{i}$ , we get

$x_{j}y_{i}-x_{i}y_{j}=-(x_{i}y_{j}-x_{j}y_{i}),$

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

$\left(\sum\limits_{i=1}^{n}x_{i}y_{i}\right)^{2}+\sum\limits_{1\leq i<j\leq n}% (x_{i}y_{j}-x_{j}y_{i})^{2}=\left(\sum_{i=1}^{n}x_{i}^{2}\right)\left(\sum_{i=% 1}^{n}y_{i}^{2}\right).$

(4)

This is equivalent to the stated identity. ∎

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