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

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 ?

Thanks

• That's only true if the $A_k$ are pairwise orthogonal. – Lord Shark the Unknown Apr 30 '18 at 17:58
• But why the equality? why the $\sum$ disappear? – user486983 Apr 30 '18 at 18:03
• $(X_n,X_m)=0$ whenever $n\ne m$. – Lord Shark the Unknown Apr 30 '18 at 18:04

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, $$\sum_{n}A_{n}(X_{n}, X_{m}) = 0 + 0 + \ldots + 0 + A_{m}(X_{m}, X_{m}) + 0 +\ldots = A_{m}(X_{m}, X_{m}).$$
Because $(X_n,X_m)=0$ if $n\ne m$. That's the orthogonality.
• But I don't see how that help to understand why $\sum_nA_n(X_n,X_m)=A_m(X_m,X_m)$ – user486983 Apr 30 '18 at 18:17
• Not sure what you don't understand. All the terms with $n\ne m$ are zero. – Martin Argerami Apr 30 '18 at 18:20