LagrangesIdentity 2002-12-21 17:16:32 2007-02-25 10:50:47 Theorem % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions \usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here Let $R$ be a commutative ring, and let $x_1, \ldots, x_n, y_1, \ldots, y_n$ be arbitrary elements in $R$. Then \[\left(\sum_{k=1}^n x_ky_k\right)^2 =\left(\sum_{k=1}^n x_k^2\right)\left(\sum_{k=1}^n y_k^2\right) - \sum_{1 \le k < i \le n} (x_ky_i -x_iy_k)^2\mbox{.}\] \begin{proof} Since $R$ is commutative, we can apply the binomial formula.We start out with \begin{equation} \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 \end{equation} Using the binomial formula, we see that \[(x_iy_j -x_jy_i)^2 =x_i^2y_j^2 -2x_ix_jy_iy_j +x_j^2y_i^2.\] So we get \begin{eqnarray} \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 &=&\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) \\ &=&\left(\sum\limits_{i=1}^n x_i^2\right)\left(\sum\limits_{i=1}^n y_i^2\right) \end{eqnarray} Note that changing the roles of $i$ and $j$ in $x_iy_j -x_jy_i$, we get \[x_jy_i -x_iy_j =-(x_iy_j -x_jy_i),\] but the negative sign will disappear when we square. So we can rewrite the last equation to \begin{equation} \left(\sum\limits_{i=1}^n x_iy_i\right)^2 +\sum\limits_{1 \le i <j \le n} (x_iy_j -x_jy_i)^2 =\left(\sum_{i=1}^n x_i^2\right)\left(\sum_{i=1}^n y_i^2\right). \end{equation} This is equivalent to the stated identity. \end{proof}