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Let $\{x_n\}_{n=1}^N$ be an orthonormal set in an inner product space $V$ with inner product $(\cdot,\cdot)$.

I am trying to show that $$\left(\sum\limits_{n=1}^N(x_n,x)x_n,x-\sum\limits_{n=1}^N(x_n,x)x_n\right)=0.$$

Reed & Simon, Functional Analysis, Theorem II.1 mention this is a “short computation using the properties of inner products” but I am having trouble with this.

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I'm guessing you meant $\,(x_n,x)\,$ in the first sum...well, in fact I'm not sure: that doesn't make sense. –  DonAntonio Dec 9 '12 at 13:21
    
Oh, I think I understand now: that huge parentheses is the inner product, right? –  DonAntonio Dec 9 '12 at 13:24
    
Right. $x$ is an element of $V$. –  dave Dec 9 '12 at 13:25

1 Answer 1

up vote 1 down vote accepted

$$\left(\sum\limits_{n=1}^N(x_n,x)x_n\,\,,\,\,x-\sum\limits_{n=1}^N(x_n,x)x_n\right)=\sum_{n=1}^N(x_n,x)^2-\sum_{n=1}^N\sum_{k=1}^N(x_n,x)(x_k,x)(x_n,x_k)=$$

$$=\sum_{n=1}^N(x_n,x)^2-\sum_{n=1}^N(x_n,x)^2=0$$

since $\,(x_n,x_k)=\delta_{n,k}\,$

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