simplifying the sum $\sum_{jSuppose $T$ is symmetric such that $T_{ab} = T_{ba}$ for all its indices $a,b \in \{ 1, \ldots, n \}$. If I consider the sum:
$$
\sum_{a<b} T_{ab}
$$
How can I simplify this? I $think$ that this should be equal to the following:
$$
\sum_{a<b} T_{ab} = \sum_{b=1}^{n} \sum_{a=1}^{b-1} T_{ab} = \frac{1}{2} \sum_{a,b=1}^{n} T_{ab} - \frac{1}{2}\sum_{c=1}^{n} T_{cc}
$$
However, beyond handwaving, I don't know how I could prove this explicitly. I don't trust it 100%.
 A: Let $T_{ij}$ be entries of a symmetric $n \times n$-matrix $T$ and consider the sum over all entries. Some sets naturally come into play: Let 
$$
\unicode{0x25FA} :=\{ (i,j) \colon 1 \leq j < i \leq n\} \quad\text{ and } \quad \unicode{0x25F9} :=\{ (i,j) \colon 1 \leq i < j \leq n\}
$$ 
denote the sets of all pairs of indices strictly below and strictly above of the diagonal and let 
$$
\square := \{ (i,j) \colon 1\leq i,j \leq n\}
$$ 
the set of all indices. Then, you have
$$
\sum_{(i,j) \in \,\square} T_{ij}
=\sum_{(i,j) \in\, \unicode{0x25FA}} T_{ij}  + \sum_{c = 1}^n T_{cc} + \sum_{(i,j) \in\, \unicode{0x25F9}} T_{ij} 
=\sum_{c = 1}^n T_{cc} + 2 \sum_{(i,j) \in\, \unicode{0x25FA}} T_{ij} ,
$$
where the last equality is due to the symmetry of the matrix.
Now, rewrite more "formally" (actually, the above is already formal)
$$\sum_{i,j=1}^n T_{ij} = \sum_{c = 1}^n T_{cc} + 2 \sum_{i = 1}^n \sum_{j = 1}^{i - 1} T_{ij}.
$$
This gives you $$\sum_{i = 1}^n\sum_{j = 1}^{i - 1} T_{ij} = \frac{1}{2} \left( \sum_{i,j = 1}^n T_{ij} - \sum_{c = 1}^n T_{cc}\right).$$
Note that I used the convention that an empty sum like $\sum_{i = 2}^1 x_i$ evaluates to zero. If you don't like this convention, the last equality should read $$\sum_{i = 2}^n\sum_{j = 1}^{i - 1} T_{ij} = \frac{1}{2} \left( \sum_{i,j = 1}^n T_{ij} - \sum_{c = 1}^n T_{cc}\right).$$
