Understanding double sums without matrices

Question

In our maths class we are currently studying sums, as an exercice we were given the task of proving the following : $$\sum_{1 \le j \lt k \le n} (a_k - a_j)(b_k-b_j) =\dfrac{1}{2}\sum_{1 \le j,k \le n} (a_k - a_j)(b_k-b_j)$$ Where $a_k , b_k$ are random reals.
I did some research and found people explaining these double sums using matrices but I can't understand them.
To be specific i'm looking for a more algebraic proof of this equality and grasping the meaning behind the double sum notations and the difference between the two sums shown above. Any help would be greatly appreciated!

Answer 1

$\sum_{1\le j,k \le n} (a_k-a_j)(b_k-b_j)$ = $\sum_{1\le j \lt k \le n} (a_k-a_j)(b_k-b_j)$ + $\sum_{1\le k\lt j \le n} (a_k-a_j)(b_k-b_j)$ + $\sum_{1\le k=j \le n} (a_k-a_j)(b_k-b_j)$

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$\sum_{1\le j \lt k \le n} (a_k-a_j)(b_k-b_j)$ = $\sum_{1\le k\lt j \le n} (a_k-a_j)(b_k-b_j)$

As $(a_k-a_j)(b_k-b_j)= (a_j-a_k)(b_j-b_k)$

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$\sum_{1\le k=j \le n} (a_k-a_j)(b_k-b_j)$ = 0

Answer 2

Writing down explicitly an example for a small $n$ can help facilitate both the explanation of the notation as well as the proof.

For $n=3$,

\begin{align} \sum_{1 \le j < k \le n} (a_k - a_j)(b_k - b_j) &= (a_2 - a_1)(b_2 - b_1) + (a_3 - a_1)(b_3 - b_1) + (a_3 - a_2)(b_3 - b_2) \\ \sum_{1 \le j,k \le n} (a_k - a_j)(b_k - b_j) &= (a_1 - a_1)(b_1 - b_1) + (a_1 - a_2)(b_1 - b_2) + (a_1 - a_3)(b_1 - b_3) \\ &\qquad + (a_2 - a_1)(b_2 - b_1) + (a_2 - a_2)(b_2 - b_2) + (a_2 - a_3)(b_2 - b_3) \\ &\qquad + (a_3 - a_1)(b_3 - b_1) + (a_3 - a_2)(b_3 - b_2) + (a_3 - a_3)(b_3 - b_3) \end{align}

The difference in the two summation notations is that the first one only involves terms where $k > j$, while the second line considers all pairs $(j,k)$. Verifying the equation for $n=3$ should be fairly straightforward from here, and you can generalize to arbitrary $n$.