Timeline for Prove/disprove a linear algebra inequality
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 12, 2023 at 5:08 | comment | added | Chen Zeno | Thank you for the answer! | |
| Jun 12, 2023 at 5:03 | history | bounty ended | Chen Zeno | ||
| Jun 12, 2023 at 5:03 | vote | accept | Chen Zeno | ||
| Jun 11, 2023 at 21:47 | comment | added | user42355 | I guess, if you already have the circular motion idea, you could get a shorter proof maybe: If $a \neq \pm w$, then let $e=\frac{a+w}{\|a+w\|}$ and $f=\frac{a-w}{\|a-w\|}$. Then $e,f$ are orthogonal unit vectors. Let us take new $a'=se+tf$ and $w'=se-tf$, with $s^2+t^2 = 1$. Then $S = \sum_j |u_j^T a'| (u_j^T w') = \sum_j \pm ((u_j^T e)^2 s^2 - (u_j^T f)^2 t^2) = U s^2 + V t^2 = (U-V)s^2+V$. So if $s,t$ are nonzero, then we can move them so that $S$ does not increase, until we hit a sign change, or $s$ or $t$ becomes $0$. Then we still need to prove the case $a=w$. | |
| Jun 11, 2023 at 20:16 | comment | added | Iosif Pinelis | I see now. Very impressive! | |
| Jun 11, 2023 at 19:56 | comment | added | user42355 | What we need to check is that $\|a\|^2 = \|w\|^2 = 1$ and $\sum_j |u_j^T a| (u_j^T w) = S$ remain true. So we need $x^T G^{-1} x = y^T G^{-1} y = 1$ and $x^T D y = S$, and these are true on the part of ellipse near $(x_0, y_0)$, until we hit a sign change. Geometrically, I believe $a$ and $w$ move circularly at the same speed, in opposite directions on the circle going through $\pm a$ and $\pm w$. | |
| Jun 11, 2023 at 19:45 | comment | added | user42355 | $x = G \alpha$, $\alpha = G^{-1} x$ and $a = \sum_j \alpha_j u_j$, and similarly, $y = G \beta$, $\beta = G^{-1} y$, and $w = \sum_j \beta_j u_j$, just like in the case of $(x_0, y_0)$. When we move on the ellipse, as long as $u_j^T a \neq 0$ and $u_j^T w \neq 0$ (so there are no sign changes), $D$ remains the same, and $S$ remains the same too. At the point where some $u_j^T a$ or $u_j^T w$ becomes $0$, we can use induction, and we are done. If we would go beyond this (we don't need to), then $D$ could change, so then $S$ could possibly change too (but this is not important for the proof). | |
| Jun 11, 2023 at 19:11 | comment | added | Iosif Pinelis | Can you explain why and how each point $(x,y)$ on the ellipse corresponds to $(a,w)$? Also, did you use in the last paragraph of your answer the fact $D_{j,j}=\pm1$ for all $j$? | |
| Jun 11, 2023 at 10:22 | history | answered | user42355 | CC BY-SA 4.0 |