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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

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

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  • $\begingroup$ ah now it's clear, thank you! $\endgroup$ – user486983 Apr 30 '18 at 18:32
  • $\begingroup$ why didn't I expand the sum.. $\endgroup$ – user486983 Apr 30 '18 at 18:34
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Because $ (X_n,X_m)=0$ if $n\ne m$. That's the orthogonality.

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  • $\begingroup$ I know that's the orthogonality $\endgroup$ – user486983 Apr 30 '18 at 18:16
  • $\begingroup$ But I don't see how that help to understand why $\sum_nA_n(X_n,X_m)=A_m(X_m,X_m)$ $\endgroup$ – user486983 Apr 30 '18 at 18:17
  • $\begingroup$ Not sure what you don't understand. All the terms with $n\ne m$ are zero. $\endgroup$ – Martin Argerami Apr 30 '18 at 18:20

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