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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.

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    $\begingroup$ 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 $w$? $\endgroup$ – Mike Earnest Aug 4 '18 at 16:10
  • $\begingroup$ @mike I think I see that if I make the line segmant u longer than the length of the projection will get shorter? $\endgroup$ – Jason Aug 4 '18 at 17:07
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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.

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  • $\begingroup$ 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). $\endgroup$ – Arnaud Mortier Aug 7 '18 at 0:15
  • $\begingroup$ Thanks! I was very close but made a tiny miatake which confused me. $\endgroup$ – Jason Aug 7 '18 at 5:47
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Hint: Visualize this in ${\mathbb R}^2$, projecting onto one-dimensional subspaces. The proof will generalize.

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

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