Existance of sub-vector space Prove/disprove:
For all $v\in V$ and for all $a\in\mathbb{R}$ such as $0\leq a\leq||v||$, there's a sub-vector space $U$ of $V$ such as the length of the orthogonal projection of $v$ on $U$ is equal to $a$.
Intuitivly I guess this is true (because if I imagine the board of $\mathbb{R}^2$, for every vector $v$ I can stretch a line $l$ that goes through $(0,0)$ such as $l\perp v$ and $proj_{U}(v)=a$). 
 A: If $v = 0$, there's nothing to prove. If $v \neq 0$, let's handle two cases separately:


*

*If $\dim V = 1$ then a subspace $U$ of $V$ is either $V$ or $\{ 0 \}$. In the former case, the projection of $v$ on $U$ is $v$ and in the latter the projection of $v$ on $U$ is $0$ so the result is false.

*If $\dim V > 1$ (possibly infinite), choose a unit vector $w$ which is orthogonal to $v$ and let $U = \operatorname{span} \{ (\cos \theta)w + \frac{\sin \theta}{\| v \|} v\}$ (this is a line in the plane spanned by $v,w$). Then 
$$\left \| (\cos \theta)w + (\sin \theta) \frac{v}{\| v \|} \right \|^2 = \cos^2 \theta + \sin^2 \theta = 1$$
and so 
$$\operatorname{proj}_{U}(v) = \left< v,  (\cos \theta)w + (\sin \theta) \frac{v}{\| v \|} \right> \left( (\cos \theta)w + (\sin \theta) \frac{v}{\| v \|} \right) \\= (\| v \| \sin \theta \cos \theta) w + (\| v \| \sin^2 \theta) \frac{v}{\| v \|}$$
which is of length
$$ \| \operatorname{proj}_{U}(v) \| = \|v \| | \sin \theta |.$$
By choosing $0 \leq \theta \leq \frac{\pi}{2}$, we see taht we can make the projection to be of any length between $0$ and $\| v \|$.

