4
$\begingroup$

Let consider a real vector space $\mathbb{R}^n$ of dimension $n$, where $\langle \cdot, \cdot\rangle$ and $||\cdot ||$ are the standard inner product and $\ell_2$ indeced norm.

Does there exist a transformation (a non-linear one) $T:\mathbb{R}^n\rightarrow \mathbb{R}^n$ such that it preserves orthogonality and amplify/enhance non-orthogonal cosines:

  1. For any two vectors $x,y\in\mathbb{R}^n$ it holds $\langle x,y\rangle = 0 \iff \langle T(x), T(y)\rangle = 0$, i.e. $T$ just preserves orthogonality
  2. $\frac{|\langle T(x), T(y)\rangle|}{||T(x)||\cdot||T(y)||} \gg \frac{|\langle x, y\rangle|}{||x||\cdot||y||} $, i.e. we would like to always amplify/enhance cosine between non-orthogonal vectors
$\endgroup$
6
$\begingroup$

This isn't possible, even if you let $T$ be non-linear. You can see this in $\mathbb{R}^2$ as follows, but the intuition generalizes straightforwardly for larger $n$:

Let $\{e_1,e_2\}$ be the standard basis of $\mathbb{R}^2$ and let $v := (1,1)/\sqrt{2}$. Then $\langle T(e_1), T(e_2) \rangle = 0$. You want $T(v)$ to have an angle of strictly less than $\pi/4$ with each of $T(e_1)$ and $T(e_2)$, which is impossible (angles satisfy the triangle inequality in $\mathbb{R}^2$, so (angle between $T(e_1)$ and $T(v)$) $+$ (angle between $T(v)$ and $T(e_2)$) $\geq$ (angle between $T(e_1)$ and $T(e_2)$) $= \pi/2$.

$\endgroup$
  • $\begingroup$ Hi Nathaniel, good point, thank you for answer. $\endgroup$ – Xorwell Dec 23 '13 at 11:58

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.