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How can I simplify the expression $\sum\limits_{i=1}^n\sum\limits_{j=1}^n x_i\cdot x_j$?

$x$ is a vector of numbers of length $n$, and I am trying to prove that the result of the expression above is positive for any $x$ vector. Is it equal to $\sum\limits_{i=1}^n x_i\cdot \sum\limits_{j=1}^n x_j$? If it is then my problem is solved, because $\left(\sum\limits_{i=1}^n x_i\right)^2$ is non-negative (positive or zero).

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    $\begingroup$ Can you simplify just $\sum_{j=1}^n x_i\cdot x_j$, where $i$ is fixed? $\endgroup$
    – Thomas Andrews
    Dec 7, 2014 at 20:57
  • $\begingroup$ I am trying to prove that the result of the expression abouve is positive for any x vector. $\endgroup$
    – Serhiy
    Dec 7, 2014 at 21:01
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    $\begingroup$ Does it equal to $ \sum _{i=1}^n x_{i} \cdot \sum _{j=1}^n x_{j} $ ? If it IS then my problem is solved :) Because $ (\sum _{i=1}^n x_{i} )^2 $ is always positive $\endgroup$
    – Serhiy
    Dec 7, 2014 at 21:08
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    $\begingroup$ @Сергій: Yes. Let $a=\sum_{i=1}^nx_i$. Then $$\sum_{i=1}^n\sum_{j=1}^n(x_ix_j)= \sum_{i=1}^n\left(x_i\sum_{j=1}^nx_j\right)=\sum_{i=1}^nax_i=a^2\;.$$ $\endgroup$
    – Brian M. Scott
    Dec 7, 2014 at 21:30
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    $\begingroup$ @Сергій We know that $\left(\sum\limits_{i=1}^n x_i\right)^2\ge0$ but not necessarily $\left(\sum\limits_{i=1}^n x_i\right)^2>0$. (I would be careful about distinction between the words positive and non-negative.) $\endgroup$
    – Martin Sleziak
    Dec 8, 2014 at 6:11

1 Answer 1

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Yes,

$$\sum_{i=1}^n\sum_{j=1}^nx_i x_j=\left(\sum_{i=1}^ nx_i\right)^2\;.$$

To see this, let $a=\sum_{i=1}^ nx_i$; then

$$\sum_{i=1}^n\sum_{j=1}^nx_i x_j=\sum_{i=1}^n\left(x_i\sum_{j=1}^nx_j\right)=\sum_{i=1}^na x_i=a^2\;.$$

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