$$\left(\sum_{i=1}^3 \sum_{j=1}^3 \delta_{ij}\right)\left(\sum_{m=1}^3 \sum_{n=1}^3 \delta_{mn}\right)$$ eq. 1. Should be:
I am just confused because I am getting $3 \times 3$.
I was reading a book. It it stated that
$$\left(\sum_{i=1}^3 \sum_{j=1}^3 \delta_{ij}\right)\left(\sum_{m=1}^3 \sum_{n=1}^3 \delta_{mn}\right)=81$$. Any advice? I am getting $3 \times 3$ Maybe the book is wrong. Just want to confirm.
Thanks in advance!
We just need to show that $$\sum_{i=1}^3 \sum_{j=1}^3 \delta_{ij} = 3.$$ Expanding the summation terms, we get \begin{align*} \sum_{i=1}^3 (\delta_{i1} + \delta_{i2} + \delta_{i3}) &= (\delta_{11} + \delta_{12} + \delta_{13}) + (\delta_{21} + \delta_{22} + \delta_{23}) + (\delta_{31} + \delta_{32} + \delta_{33}) \\ &= (1 + 0 + 0) + (0 + 1 + 0) + (0 + 0 + 1) = 3. \end{align*}