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I wish to find a proof for this equality:

$(a_1 + a_2 + a_3)^2 = a_1^2 + a_2^2 + a_3^2 + 2(a_1a_2+ a_2a_3 + a_1a_3)$

But then I realized there exists a more general version:

$(\sum\limits_{k=1}^n a_k)^2 = \sum\limits_{k=1}^n a_k^2 + 2 \sum\limits_{1 \leq k < j \leq m} a_ka_j$

Does anyone have a proof for the more general version of this equality?

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    $\begingroup$ I think proving by induction would be straightforward. Also in the last summation sign it should be $n$ instead of $m$. $\endgroup$ – Empiricist Feb 2 '16 at 4:31
  • $\begingroup$ Why not use binomial twice $\endgroup$ – Archis Welankar Feb 2 '16 at 5:40
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Lemma: Lagrange identity Let $x=(x_1,...,x_n),y=(y_1,...,y_n)\in\mathbb{R}^n$. Then $\|x\|^2 \|y\|^2-\langle x,y\rangle^2=\sum_{i<j}(x_iy_j-x_jy_i)^2$.

Before prove this formula, note that it means

$$\sum_{i=1}^nx_i^2\sum_{j=1}^ny_i^2=\sum_{i<j}(x_iy_j-x_jy_i)^2+\left(\sum_{i=1}^nx_iy_i\right)^2.$$

Thus, if we take $y_i=1$, then

$$n\sum_{i=1}^nx_i^2=\sum_{i<j}(x_i-x_j)^2+\left(\sum_{i=1}^nx_i\right)^2\\ =\sum_{i<j}(x_i^2+x_j^2)-2\sum_{i<j}x_ix_j+\left(\sum_{i=1}^nx_i\right)^2\\=(n-1)\sum_{i=1}^nx_i^2-2\sum_{i<j}x_ix_j+\left(\sum_{i=1}^nx_i\right)^2$$ from your equality follows.

Proof of Lagrange Identity

$\begin{eqnarray} &&\sum_{i=1}^nx_i^2\sum_{j=1}^ny_i^2-\left(\sum_{i=1}^nx_iy_i\right)^2\\ &=&\sum_{i=1}^nx_i^2\sum_{j=1}^ny_i^2-\sum_{i=1}^nx_iy_i\sum_{j=1}^nx_jy_j\\ &=&\sum_{i,j=1}x_i^2y_j^2-\sum_{i,j=1}x_iy_ix_jy_j\\ &=&\sum_{i<j}x_i^2y_j^2+\sum_{i=1}x_i^2y_i^2+\sum_{i>j}x_j^2y_i^2\\ &&-\sum_{i<j}x_iy_ix_jy_j-\sum_{i=j}x_iy_ix_jy_j-\sum_{i>j}x_iy_ix_jy_j\\ &=&\sum_{i<j}x_i^2y_j^2+\sum_{i>j}x_j^2y_i^2-2\sum_{i<j}x_iy_ix_jy_j\\ &=&\sum_{i<j}x_i^2y_j^2-2x_iy_ix_jy_j+x_jy_i^2\\ &=&\sum_{i<j}(x_iy_j-x_jy_i)^2 \end{eqnarray}$

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  • $\begingroup$ This seems like heavy overkill. A simple inductive proof would do $\endgroup$ – ASKASK Feb 2 '16 at 5:23
  • $\begingroup$ Indeed there is a simpler proof by simmetry. On the other hand, if you know Lagrange Indentity, then the OP is trivial as I show $\endgroup$ – sinbadh Feb 2 '16 at 5:25
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Both describe the quadratic form defined by $A^T G A,$ where the column vector $$ A = \left( \begin{array}{c} a_1 \\ a_2 \\ \vdots \\ a_n, \end{array} \right) $$ and $G$ is the square matrix with all entries equal to $1.$

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