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To understand a proof in functional analysis I need to understand why the following equation is true:

$$\lVert x\rVert^2 - \sum_{j=1}^n |x_i|^2 = \Biggl\lVert x-\sum_{i=1}^nx_ie_i\Biggr\rVert^2$$

Where $x\in H$ ($H$ a Hilbert space) and $x_i= \langle x,e_i\rangle$ and $e_i$ is an orthonormal system.

Can someone explain me why this equality is true?

Thanks in advance!

Daniel Fischer
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Duke
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    The right hand side is $$\Biggl\langle x - \sum_{i = 1}^n x_i e_i , x - \sum_{i = 1}^n x_i e_i \Biggr\rangle.$$ Expand that using bilinearity or sesquilinearity, and simplify using $x_i = \langle x, e_i\rangle$. – Daniel Fischer Jun 30 '15 at 22:09
  • I got $||x||^2-2Re(<x,\sum_{i=1}^nx_ie_i>)+\sum_{j=1}^n|x_i|^2$. Is this correct? How can I rewrite this to get the LHS? – Duke Jun 30 '15 at 22:17
  • @Duke: $\langle x_i,e_i\rangle$ makes no sense, since $x_i$ is a number. – Martin Argerami Jun 30 '15 at 22:19
  • @MartinArgerami Edited it :) – Duke Jun 30 '15 at 22:20
  • Well, now calculate the term inside the brackets (there 's a bracket missing, by the way). Then you'll be done and you can write an answer to your own question :) – Martin Argerami Jun 30 '15 at 22:21
  • The "Re" is fine, until he calculates the inner product :) – Martin Argerami Jun 30 '15 at 22:29
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    The $x_i$ are chosen so that $x-\sum_{i=1}^{n}x_i e_i$ is orthogonal to every element of ${ e_1,e_2,\cdots,e_n}$. So this is an orthogonal decomposition: $$ x = \left[x-\sum_{i=1}^{n}x_i e_i\right]+\sum_{i=1}^{n}x_i e_i. $$ Therefore, by the Pythagorean identity: \begin{align} |x|^{2} & =\left|x-\sum_{i=1}^{n}x_i e_i\right|^{2}+\left|\sum_{i=1}^{n}x_i e_i\right|^{2} \ & = \left|x-\sum_{i=1}^{n}x_i e_i\right|^{2}+\sum_{i=1}^{n}|x_i|^{2}. \end{align} – Disintegrating By Parts Jun 30 '15 at 22:58

3 Answers3

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\begin{align} \left\|x-\sum \limits_{i=1}^{n}\langle x,e_i\rangle e_i\right\|^2&=\langle x-\sum \limits_{i=1}^{n}\langle x,e_i\rangle e_i\,,x-\sum \limits_{i=1}^{n}\langle x,e_i\rangle e_i\ \rangle \\ &=\langle x,x\rangle+\underbrace{\langle \sum \limits_{i=1}^{n}\langle x,e_i\rangle e_i,\sum \limits_{i=1}^{n}\langle x,e_i\rangle e_i\rangle}_{\text{(by Pythagoras Theorem)}}-\langle x,\sum \limits_{i=1}^{n}\langle x,e_i\rangle e_i\rangle-\langle \sum \limits_{i=1}^{n}\langle x,e_i\rangle e_i,x\rangle \hspace{-10 mm} \\ &=\|x\|^2+\sum\limits_{i=1}^{n}|\langle x,e_i\rangle|^2\|e_i\|^2-\sum \limits_{i=1}^{n}\overline{\langle x,e_i\rangle}\langle \overline{e_i},\overline{x}\rangle-\sum \limits_{i=1}^{n}\langle x,e_i\rangle\langle e_i,x\rangle \\ &=\|x\|^2+\sum\limits_{i=1}^{n}|\langle x,e_i\rangle|^2-\sum\limits_{i=1}^{n}|\langle x,e_i\rangle|^2-\sum\limits_{i=1}^{n}|\langle x,e_i\rangle|^2 \\ &=\|x\|^2-\sum\limits_{i=1}^{n}|\langle x,e_i\rangle|^2 \end{align}

Eugene Zhang
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For $1\le j\le n$, $$ \begin{align} \left\langle x-\sum_{k=1}^n\langle x,e_k\rangle e_k,x\right\rangle &=\langle x,x\rangle-\sum_{k=1}^n\langle x,e_k\rangle\langle x,e_k\rangle\\ &=\|x\|^2-\sum_{k=1}^n\langle x,e_k\rangle^2\tag{1} \end{align} $$ and $$ \begin{align} \left\langle x-\sum_{k=1}^n\langle x,e_k\rangle e_k,e_j\right\rangle &=\langle x,e_j\rangle-\sum_{k=1}^n\langle x,e_k\rangle\overbrace{\langle e_k,e_j\rangle}^{\delta_{j,k}}\\ &=\langle x,e_j\rangle-\langle x,e_j\rangle\\[9pt] &=0\tag{2} \end{align} $$ Putting together $(1)$ and $(2)$ yields $$ \begin{align} \left\|x-\sum_{k=1}^n\langle x,e_k\rangle e_k\right\|^2 &=\left\langle x-\sum_{k=1}^n\langle x,e_k\rangle e_k,x-\sum_{k=1}^n\langle x,e_k\rangle e_k\right\rangle\\[6pt] &=\|x\|^2-\sum_{k=1}^n\langle x,e_j\rangle^2\tag{3} \end{align} $$

robjohn
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The main thing: $y = y_1 + \ldots + y_m$ and the $y_i$'s are paiwise orthogonal ( an orthogonal decomposition) then: $$||y||^2 = ||y_1||^2 + \cdots + ||y_m||^2 $$

It's easy to check that if $e_1$, $\ldots$, $e_n$ is an $\it{orthonormal}$ family then $x - \sum_{i=1}^n \langle x, e_i \rangle e_i$ is perpendicular to all the $e_i$'s and so we have for any scalars $\delta_i$ the orthogonal decomposition $$x-\sum \delta_i e_i = \sum_{i=1}^n (\langle x, e_i \rangle - \delta_i)e_i + (x - \sum_{i=1}^n \langle x, e_i \rangle e_i)$$ and so $$||x - \sum \delta_i e_i||^2 = \sum_{i=1}^n |\delta_i - \langle x, e_i \rangle|^2 + ||x - \sum_{i=1}^n \langle x, e_i \rangle e_i||^2$$

Therefore $\sum_{i=1}^n \langle x, e_i \rangle e_i$ is the (unique) closest point to $x$ in the span of $e_i$.

orangeskid
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