Show $\frac{1}{2}\sum_{i\neq j}S_{ij}=\sum_{i<j}S_{ij}$

Question

I am looking to prove the following $$ \frac{1}{2}\sum_{i\neq j}S_{ij}=\sum_{i<j} S_{ij},\qquad S_{ij}=S_{ji}. $$ I am not sure how to understand why it works. Thanks

Answer 1

Hint:

$$ \newcommand{\t}{\times} \begin{array}{cccccc} {} & 1 & 2 & 3 & 4 & 5 \\ 1 & {\Large\circ} & \t & \t & \t & \t \\ 2 & \t & {\Large\circ} & \t & \t & \t \\ 3 & \t & \t & {\Large\circ} & \t & \t \\ 4 & \t & \t & \t & {\Large\circ} & \t \\ 5 & \t & \t & \t & \t & {\Large\circ} \end{array} $$

Answer 2

This is only true if the things $a_{ij}$ being summed over (missing in your question) are symmetric with respect to $i$ and $j$, which is to say $a_{ji} = a_{ij}$.

For example, if $1 \le i, j \le 3$, then $\sum_{i\ne j} a_{ij} = a_{12} + a_{13} + a_{21} + a_{23} + a_{31} + a_{32}$, whereas $\sum_{i < j} a_{ij} = a_{12} + a_{13} + a_{23}$ and $\sum_{i > j} a_{ij} = a_{21} + a_{31} + a_{32}$.

Now, if we have symmetry, then $a_{21} + a_{31} + a_{32} = a_{12} + a_{13} + a_{23}$. Does this look familiar?

Answer 3

Since $S_{ji}=S_{ij}$, and if $i\ne j$, then either $i\lt j$ or $j\lt i$ $$ \begin{align} \sum_{i\ne j}S_{ij} &=\sum_{i\lt j}(S_{ij}+S_{ji})\\ &=2\sum_{i\lt j}S_{ij} \end{align} $$