Vtyjkyuk

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
$$sigma(v) = sum_i=1^n langle v,w_irangle w_i$$
Show that $|sigma(v) - v |< | w - v | $ for all $vin V$ and for all $w in W$, with $w neq sigma(v)$.

Take an arbitrary $vin V$ such that $sigma(v)neq w$ and then,
beginalign*
|sigma(v) - v | &= biggl|sum_i=1^n langle v,w_irangle w_i - v biggr| \
&leq sum_i=1^n | langle v,w_irangle w_i - v|\
endalign*
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 $langle w_i , w_jrangle = 1$ 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.

The key observation is that $sigma(v)-v,perp,W$.

Indeed, $langlesigma(v),w_irangle=langle v,w_irangle$ 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:

beginalign*
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_irangle 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.
endalign*

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.