3
$\begingroup$

I have unit balls defined by the $1$, $2$ and $\infty$ norm in $\mathbb{R}^2$. I want to find the orthogonal projection of a vector $(x,y)$ onto the balls.

How could it be done? I only know how to project vectors orthogonally but have no idea how to do it over the balls.

Thanks a lot.

| cite | improve this question | | | | |
$\endgroup$
  • 1
    $\begingroup$ I'm not sure what an "orthogonal projection over a ball" means. You usually project over a vector space (geometrically, a line, a plane, etc). Perhaps we mean a non-orthogonal projection, just the scaled vector that lies on the ball. $\endgroup$ – leonbloy May 5 '12 at 23:10
  • $\begingroup$ @leonbloy, I mean the orthogonal projection, if you look at the 2-norm in $\mathbb{R}^2$, it is a circle. and for $\infty$-norm it is a square... I hope it clarifies. $\endgroup$ – Shan May 6 '12 at 10:07
  • $\begingroup$ not really for me $\endgroup$ – leonbloy May 6 '12 at 12:34
  • 1
    $\begingroup$ I think Shan calls "orthogonal projection" the point on the sphere which is closest to the given vector. $\endgroup$ – Siminore May 18 '12 at 9:08
  • $\begingroup$ @Siminore, Yup this is correct! $\endgroup$ – Shan May 21 '12 at 9:53
1
$\begingroup$

Recall that the orthogonal projection of a vector $u$ on a vector subspace $V$ is defined as the unique vector $v$ such that (1) $v$ is in $V$, and (2) $u-v$ is orthogonal to $V$, that is, $\langle u,w\rangle=\langle v,w\rangle$ for every $w$ in $V$.

What could be the orthogonal projection of $u$ on a sphere $S$? (In the question you write ball but in the comments it seems clear you mean the unit sphere for a given norm.) Following the classical definition, one should look for $s$ such that (1) $s$ is in $S$ and (2) $u-s$ is orthogonal to... what exactly? Orthogonal to $S$? Alas, no vector $r$ is orthogonal to $S$ in the sense that $\langle r,t\rangle=0$ for every $t$ in $S$, except the null vector. (For example, $t=\langle r,r\rangle^{-1/2}r$ is in $S$ and $\langle r,t\rangle=\langle r,r\rangle^{1/2}\ne0$ for every $r\ne0$.)

This remark shows that you really need to clarify the question.

| cite | improve this answer | | | | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.