# A double summation

While going through the article given here (page 23), it seems that the following doublesummation is carried out

$$\sum_{i=1,j=1}^{N^2} B_{ij} = \Big( \sum_{i,j=1}^{N^2-1} + \sum_{i=1,j=N^2}^{N^2} + \sum_{i=N^2,j=1}^{N^2} - \sum_{i=N^2,j=N^2}^{N^2} \Big) B_{ij}$$. Where $$B_{ij} = c_{ij} F_i \rho_S F_j^\dagger$$, is an operator.

Could anyone help me to understand how the double sum on the left breaks into the four sums on the right hand side?

When you write $$\sum_{i,j=1}^{N^2-1} B_{ij}$$, to get to $$\sum_{i,j}^{N^2}B_{ij}$$ you are missing all the terms that have either $$i=N^2$$, or $$j=N^2$$. The authors are dividing these in two groups:
• $$i=1,\ldots,N^2$$; $$j=N^2$$
• $$i=N^2$$; $$j=1,\ldots,N^2$$
But now they are counting $$i=N^2$$, $$j=N^2$$ twice, so they subtract it once.