Show that a subspace $U$ exists such that the orthogonal projection of $v$ onto it has the specified length $\alpha < \|v\|$

Question

Given the inner product space $V$ such that $\dim V > 1$.

Show that for every $\alpha \in \mathbb R$ and $v \in V$ such that $0\le \alpha \le\| v\|$ there exists a subspace $U\subset V$ such that the orthogonal projection of $v$ onto $U$ has length $\alpha$.

I think this has something to do with representing $U$ in an orthonormal basis $\{u_1\}$ for example and then saying $P_U(v)=\langle v,u_1\rangle v$. But I'm not really sure about this.

Answer 1

As in my comment, let $w$ be a nonzero vector which is perpendicular to $w$. Such a $w$ exists; take any nonzero $x$ which is not a scalar multiple of $v$ (which exists since $\dim V>1$), and let $w=x-\frac{\langle{x,v}\rangle}{\langle{v,v}\rangle}v$.

Recall that for any $u\neq 0$, the length of the projection of $v$ onto the subspace generated by $u$ is $$ \frac{|\langle u,v\rangle |}{\|u\|} $$ Now, let $u=tw+(1-t)v$. Using the above formula, the length of the projection is $$ \frac{|\langle tw+(1-t)v,v\rangle |}{\|tw+(1-t)v\|}=\frac{|\langle tw+(1-t)v,v\rangle |}{\sqrt{\langle tw+(1-t)v,tw+(1-t)v\rangle}}=\frac{|1-t|\cdot \|v\|^2}{\sqrt{t^2\|w\|+(1-t)^2\|v\|^2}} $$ Now, consider the above as a function of $t$. We notice three things:

• When $t=0$, the function is $\|v\|$.
• When $t=1$, the function is $0$.
• The function is continuous in $t$.

By the intermediate value theorem, for every $0<\alpha<\|v\|$, there exists a $t$ for which the value of the function (the length of the projection) is equal to $\alpha$, as desired.

Answer 2

Hint: Visualize this in ${\mathbb R}^2$, projecting onto one-dimensional subspaces. The proof will generalize.