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

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.

• Let $w$ be a vector perpendicular to $v$. When you project $v$ onto $\text{span}\{w\}$, the length is 0. When you project $v$ onto $\text{span}\{v\}$, the length is $\|v\|$. What happens when you project onto $\text{span}\{u\}$, where $u$ is on the line segment connecting $v$ to $w$? – Mike Earnest Aug 4 '18 at 16:10
• @mike I think I see that if I make the line segmant u longer than the length of the projection will get shorter? – Jason Aug 4 '18 at 17:07

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.

• For completeness you could add how to find a perpendicular vector (which is where the assumption that the dimension is more than one is necessary). – Arnaud Mortier Aug 7 '18 at 0:15
• Thanks! I was very close but made a tiny miatake which confused me. – Jason Aug 7 '18 at 5:47

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

• I think I can visualize it but I'm having troible writing down a proof – Jason Aug 4 '18 at 17:22