Timeline for Does the Perron vector maximize $x^TAx$ in the simplex?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 19, 2019 at 0:46 | vote | accept | dineshdileep | ||
| Sep 18, 2019 at 3:08 | comment | added | dineshdileep | @AnthonyQuas got it!... | |
| Sep 17, 2019 at 23:36 | comment | added | Anthony Quas | The spirit of the answer is that $\ell^2$ looks very different from $\ell^1$ in the diagonal direction (by a factor of $\sqrt n$). Finding a matrix whose Perron vector is close to the diagonal direction means that the $\ell^1$ version of it is much smaller. This makes it easier to beat the value coming from the Perron vector. As an explicit perturbation along the lines suggested by @JochenGlueck, you could replace all of the 1/3 by 1/4; and replace the 0’s by $1/(4(n-3))$. The Perron vector is the same so $x^TAx=1/n$ and $y^TAy=1/4$. | |
| Sep 17, 2019 at 6:26 | comment | added | Jochen Glueck | @dineshdileep: No, this doesn't make a difference. You can simply perturb the matrix to make all entries strictly positive, but if the perturbation is small enough, this won't change the phenomenon described in Anthony Quas' answer. | |
| Sep 17, 2019 at 3:47 | comment | added | dineshdileep | Thanks, this is insightful. But please note that the matrix I have is strictly positive ($A_{ij}>0$) whereas the matrix in your example is extremely sparse for larger $n$. Do you think this will make a difference? | |
| Sep 17, 2019 at 1:28 | history | answered | Anthony Quas | CC BY-SA 4.0 |