Timeline for Rescaling positive definite matrices to force a unit eigenvector
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 15, 2013 at 1:38 | answer | added | David Bryant | timeline score: 3 | |
| Jan 13, 2013 at 8:36 | vote | accept | David Bryant | ||
| Jan 13, 2013 at 6:13 | answer | added | Will Sawin | timeline score: 4 | |
| Jan 13, 2013 at 4:22 | comment | added | Will Sawin | I believe I have a proof for $X$ nonnegative. Consider the simplex of diagonal matrices $W$ with nonnegative entries up to scaling, and the simplex of vectors $V$ with nonnegative entries up to scaling. The map $V= WX'XW$ sends each $k$-cell of the first simplex into the corresponding $k$-cell of the second simplex, since if some of the coordinates of $W$ are $0$ then some of the coordinates of $V$ are $0$. Every such map on simplices must be surjective, because it has a boundary-preserving homotopy to the standard one, by induction. | |
| Jan 13, 2013 at 3:04 | comment | added | Brendan McKay | There's a theorem that indecomposible nonnegative matrices can be rescaled (using separate row and column scalings) to become doubly stochastic. The proof is not trivial. I don't know if that can help, but it feels similar. | |
| Jan 12, 2013 at 22:58 | comment | added | David Bryant | @MTS: I'm assuming real-valued entries, thx. Also, by symmetry, we will also have that WX'XW has columns that sum to one. It can have negative entries, though, so need not be doubly-stochastic. | |
| Jan 12, 2013 at 22:56 | history | edited | David Bryant | CC BY-SA 3.0 |
added 99 characters in body
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| Jan 12, 2013 at 22:14 | comment | added | Alexandre Eremenko | That this is true for 2 times 2 matrices :-) | |
| Jan 12, 2013 at 22:12 | comment | added | MTS | You probably already know this, but the vector of all ones is an eigenvector of a matrix $M$ if and only if the rows of $M$ all have the same sum. So you want to try to find your diagonal matrix $W$ so that the row sums of $WX'XW$ are all one. (Also, you should make it clear what you mean by $X'$; is this the transpose or the conjugate transpose, i.e. are you working over $\mathbb{R}$ or $\mathbb{C}$?) | |
| Jan 12, 2013 at 21:26 | history | asked | David Bryant | CC BY-SA 3.0 |