# Understanding double sums without matrices

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!

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

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

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

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

• Exactly what I needed. Thank you! – OUCHNA Sep 20 '20 at 18:23

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 $$n$$.

• Thanks for your input! – OUCHNA Sep 20 '20 at 18:24