multiplication of finite sum (inner product space) I am having difficulty to understand the first line of the proof of theorem 3.22 below. (taken from a linear analysis book)

Why need to be different index, i.e. $m,n$ when multiplying the two sums? This is very basic, but I really need help for explanations.

Orthornormality refers to the basis $e_i$. When a basis is orthonormal it means the inner product between any two elements of the basis $e_i,e_j$ is $\langle e_i, e_j \rangle = \delta_{ij}$ (see kronecker delta). More generally, two vectors $u,v$ are orthogonal if $\langle u, v \rangle = 0$. The normality part comes from elements of the basis having norm $1$.
The part on algebraic properties refers simply to the bilinearity. If $u,v,w$ are vectors and $\alpha, \beta \in \mathbb{R}$, then $\langle \alpha \cdot u,  v\rangle=\alpha \cdot \langle u, v \rangle = \langle u, \alpha \cdot v \rangle$ and $\langle u + v, w\rangle = \langle u, w\rangle + \langle v, w\rangle$ and $\langle u,v+w\rangle = \langle u, v\rangle+\langle u, w\rangle$.
The proof uses the fact that $\lVert v \rVert^2 = \langle v, v \rangle$ and the properties above to expand the inner product, then the orthonormality to simplify it.
In the end, this theorem says that for orthonormal bases, you can use Pythagoras's Theorem to calculate lengths.
