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Let $V$ be an inner product space, and let $W$ be its finite dimensional linear subspace that has an orthonormal basis $\{ w_1, \dots , w_n \}$. Define $\sigma : V \to W$ by $$\sigma(v) = \sum_{i=1}^n \langle v,w_i\rangle w_i$$ Show that $\|\sigma(v) - v \|< \| w - v \|\ $ for all $v\in V$ and for all $w \in W$, with $w \neq \sigma(v)$.

Take an arbitrary $v\in V$ such that $\sigma(v)\neq w$ and then, \begin{align*} \|\sigma(v) - v \| &= \biggl\|\sum_{i=1}^n \langle v,w_i\rangle w_i - v \biggr\| \\ &\leq \sum_{i=1}^n \| \langle v,w_i\rangle w_i - v\|\\ \end{align*} I don't know how I utilize the fact that $\{ w_1, \dots , w_n \}$ is an orthonormal basis for $W$. I know that means that $\langle w_i , w_j\rangle = 1$ if $i=j$ and $0$ otherwise. Also, $\sigma(v) \subseteq W$, so we should be able to represent it as $\sigma(v) = (\alpha_1 w_1 , \alpha_2 w_2 , \dots , \alpha_n w_n)$ where $\alpha_i$ are scalars. Any help would be appreciated.

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The key observation is that $\sigma(v)-v\,\perp\,W$.
Indeed, $\langle\sigma(v),w_i\rangle=\langle v,w_i\rangle$ for each basis element $w_i$ of $W$.

Now, $w-v=(w-\sigma(v))\,+\,(\sigma(v)-v)$, and these two summands are orthogonal to each other by the above, as $w-\sigma(v)\in W$, so we have $$\|w-v\|^2=\|w-\sigma(v)\|^2\,+\,\|\sigma(v)-v\|^2\,.$$


On why $\sigma(v) - v \perp W$. This means that $\langle \sigma(v) - v , w_i \rangle = 0$ for all $1 \leq i \leq n$. Then note that:

\begin{align*} \sum_{i=1}^n \langle \sigma(v) - v , w_i \rangle &= \sum_{i=1}^n \langle \sigma(v) , w_i \rangle - \sum_{i=1}^n \langle v , w_i \rangle \\ &= \sum_{i=1}^n \langle \sum_{j=1}^n \langle v, w_j \rangle w_j , w_i \rangle - \sum_{i=1}^n \langle v,w_i \rangle \\ &= \sum_{i=1}^n \langle \langle v,w_i\rangle w_i , w_i \rangle- \sum_{i=1}^n \langle v, w_i \rangle \\ &= \sum_{i=1}^n \langle v,w_i \rangle - \sum_{i=1}^n \langle v,w_i \rangle \\ &=0. \end{align*}

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  • $\begingroup$ Why is $\sigma(v) - v \perp W$, what in the question gives that away? I must be missing a foundational concept. Given that I see how the rest of the argument is formed. Thank you! $\endgroup$ – Dragonite Sep 11 '18 at 12:45
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You can write any vector of V as the sum of a vector that lies in W and one that is orthogonal to W. Geometrically, you're trying to show that the orthogonal projection of v onto W is the vector in W closest to v. Think of the vector v as one leg of a triangle and start connecting v to vectors w in W. You should be able to at least convince yourself that the shortest distance from w to v is when you have a right triangle. The argument in the previous comment proves it.

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