Proving that $||v - \text{proj}_F^\perp (v)|| \le ||v - u||,\forall u \in F $

Question

I'm having some trouble proving the following proposition:

Let $V$ be a finite-dimensional inner product real vector space: For each $v\in V$ it's orthogonal projection over $F \le V$ (denoted by: $\text{proj}_F^\perp (v)$) is the closest vector of $F$ to $v$, in other words: $$||v - \text{proj}_F^\perp (v)|| \le ||v - u||,\forall u \in F $$

I tried a direct proof, I tried a proof by contradiction, but I wan's able to prove this proposition. How can this proof be done?

Answer 1

If $v\in V$, and if $f\in F$ and $f'\in F^\perp$ are such that $v=f+f'$, then $f=\operatorname{proj}F^\perp(v)$ and, for any $u\in F$,\begin{align}\|v-u\|^2&=\bigl\|(v-f)+(f-u)\bigr\|^2\\&=\|v-f\|^2+\|f-u\|^2\text{ (because $f-u\in F$ and $v-f\in F^\perp$)}\\&\geqslant\|v-f\|^2.\end{align}