0
$\begingroup$

Let $x$ be a vector and $A$ a matrix. Let $$ y := \dfrac{Ax}{\lVert A \rVert} $$ bet a unit transform of vector $x$, and $$ \widehat{y}:= y/\lVert x\rVert $$ the same transformation with normalization on $x$. Also define a vector $z$ point-wisely by $$ z:=e^y $$ with $ \overline{z} := z/\lVert z \rVert$.

Question. Suppose $\widehat{y}_1>\widehat{y}_2$ (or equivalently $y_1 >y_2$) where $y_1$ and $y_2$ are the first and second elements of $y$, respectively. Then $$ \overline{z}_1 - \overline{z}_2 \geq \widehat{y}_1 - \widehat{y}_2? $$

The assumption that the transformation is a unit one might be removed; we might simply replace $A/\lVert A \rVert$ by $ M$. Would it then be easier to prove or disprove?

Additional Assumption. Suppose $x$ is such that $y_i >0$ for all $i$ and $y_1 = \max y_i$.

$\endgroup$
  • $\begingroup$ When you are comparing two vectors, are you comparing them element wise? $\endgroup$ – sudeep5221 May 21 at 13:05
  • $\begingroup$ @sudeep5221 Sorry for unclear writing. I will clarify. $\endgroup$ – jachilles May 21 at 13:08
1
$\begingroup$

Let $A$ be identity matrix, $x = (4a, a, 8a)$. Then $\hat y = (\frac{4}{9}, \frac{1}{9}, \frac{8}{9})$ and $\hat y_1 - \hat y_2 = \frac{1}{3}$.

But $\|z\| > e^{8a}$ so $\bar z_1 < e^{-4a}$. As $\bar z_2 > 0$ we have $\bar z_1 - \bar z_2 < e^{-4a}$, so for large enough $a$ we have $\bar z_1 - \bar z_2 < \frac{1}{3}$.

$\endgroup$
  • $\begingroup$ Thank you for your answer, sincerely. But since the derivative of $e^y$ is smaller than linear one when $y_i<0$, I think the above conjecture would not hold. And you provided a counter example. Sorry to bother but how about when $x$ is such that all the components $y_i>0$ $\endgroup$ – jachilles May 22 at 5:22
  • 1
    $\begingroup$ In my example all $y_i > 0$. $\endgroup$ – mihaild May 22 at 8:23
  • 1
    $\begingroup$ Take $x = (4 \varepsilon, 3 \varepsilon)$ for some small $\varepsilon$. Then $\hat y_1 - \hat y_2 = \frac{1}{5}$, but $\bar z_1 - \bar z_2 \approx \frac{\varepsilon}{2}$. $\endgroup$ – mihaild May 22 at 8:43
  • 1
    $\begingroup$ If we omit $A$ then it probably is, but I can't came with formal sentence. I believe it's better to be asked as separate question - much more people will notice a question then comments buried in existent question. $\endgroup$ – mihaild May 22 at 9:11
  • 1
    $\begingroup$ And of course if we don't omit $A$, lower bound on $x$ is irrelevant, as $\frac{A}{\|A\|}$ can decrease norm. $\endgroup$ – mihaild May 22 at 9:27

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.