Why $\sum_nA_n(X_n,X_m)=A_m(X_m,X_m)$?

Question

Why in the following proof $$\sum_nA_n(X_n,X_m)=A_m(X_m,X_m)$$ ?

The author says it's because orthogonality but orthogonality means $(f,g)=\int_a^bfgdx=0$. So how come orthogonality helps to prove it ?

Could someone explain this please?

Thanks

Answer 1

To elaborate on other answers/comments, observe that $$ \sum_{n}A_{n}(X_{n}, X_{m}) = A_{1}(X_{1}, X_{m}) + A_{2}(X_{2}, X_{m}) + \ldots + A_{m}(X_{m}, X_{m}) + \ldots $$ and all of the terms where the index of $A$ is not $m$ are zero. So, $$ \sum_{n}A_{n}(X_{n}, X_{m}) = 0 + 0 + \ldots + 0 + A_{m}(X_{m}, X_{m}) + 0 +\ldots = A_{m}(X_{m}, X_{m}). $$

Answer 2

Because $ (X_n,X_m)=0$ if $n\ne m$. That's the orthogonality.