Adding conditions on double sums

Question

Suppose now we have $a_{jk}$ where $1 \leq j,k \leq n$. If $a_{jk} = a_{kj}$, then we have

\begin{equation} \sum_{1 \leq j < k \leq n} a_{jk} = \frac{1}{2} \sum_{j=1}^n \sum_{k=1}^n a_{jk} \end{equation}

Is this a true statement?

Answer 1

Not in general. Think of the $a_{jk}$ as being entries in an $n\times n$ matrix $A$, with $a_{jk}$ in row $j$, column $k$. Then

$$\sum_{1\le j<k\le n}a_{jk}$$

is the sum of the elements above the main diagonal of $A$, while

$$\sum_{j=1}^n\sum_{k=1}^na_{jk}$$

is the sum of all of the elements of $A$. There’s no reason for the first of these to be half of the second in general. If $A$ is symmetric, so that $a_{jk}=a_{kj}$ for all $j,k\in\{1,\ldots,n\}$, then the sum of the elements above the diagonal is the same as the sum of the elements below the diagonal, but that still doesn’t yield your equation, unless the sum of the elements on the main diagonal is $0$.