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{.}

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.∎