Timeline for Eigenvalues of permutations of a real matrix: can they all be real?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Sep 11, 2013 at 20:17 | comment | added | Wolfgang | I forgot to say: this is valid for T=2. I wanted to avoid logarithms etc. to write it in a "purely exponential" form. | |
| Sep 11, 2013 at 20:05 | comment | added | Wolfgang | (cont'd) Here it is : $$M=\pmatrix {1&T^{17}&T^{ \color{red}{31}}&T^{45}&T^{56}&T^{65}&T^{72} \\0&1&T^{17}& T^{32}&T^{45}&T^{56}\cdot\color{red}{\frac{299}{256}}&T^{65} \\0&0&1&T^{17}& T^{32}\cdot\color{red}{\frac94}&T^{45}&T^{56} \\0&0&0&1&T^{17}&T^{32}&T^{45} \\0&0&0&0&1&T^{17}&T^{\color{red}{31}} \\0&0&0&0&0&1&T^{17} \\0&0&0&0&0&0&1}$$ | |
| Sep 11, 2013 at 20:04 | comment | added | Wolfgang | I have pursued the “kn-k(k+3)/2” thing and replaced the “3” by other, smaller numbers. The behaviour is similar, very few non-real zeros. Below -4, there is no further improvement. But I’ve found for $n=7$ that changing only 4 values results in a matrix with as little as $c(M)=4$. Almost there ! Sadly, the four remaining pairs all have $\Im/\Re\approx .6938$, quite far from zero. ... | |
| Sep 7, 2013 at 21:17 | comment | added | S. Carnahan♦ | @Wolfgang That is quite an impressive improvement! Perhaps I just found an inconvenient local minimum. | |
| Sep 7, 2013 at 20:23 | comment | added | Wolfgang | in fact, not even necessarily "breaking the symmetry", but putting 'bumps' and 'dents' on the first diagonal(s), which have so far been constant. Probably nothing 'concave' either, so the search for a generalisable pattern seems quite desperate so far. But an interesting question: why does an exponent sequence [0,n-2,2n-5,3n-9,...,kn-k(k+3)/2,...] with constant diagonals yield relatively few complex eigenvalues? | |
| Sep 7, 2013 at 19:04 | comment | added | Wolfgang | (cont'd) This can certainly be improved by breaking the symmetry for n=7. My code takes 3 min, so I left it there. (for the moment) :) For n=7 and T=4, more than half of the complex roots have |Im/Re|<.025, so there is much room for improvements. | |
| Sep 7, 2013 at 19:00 | comment | added | Wolfgang | Well, my best attempt was inspîred by your rank 5 example. I noticed that the exponents 0,3,5,6 (ignoring the last one, which can be anything from 0 to 6) have differences 3,2,1. And generalising this seems quite promising as a start: Taking 0,4,7,9,10,x for rank 6 yields a defect of 11 (and now I don't believe anymore that a solution with constant diagonals is possible for n>5). Likewise, 0,5,9,12,14,15,x for rank 7 yields only 70, much better than your 648. (Here x can be anything between 0 and 10 resp. 15). Even a slight modification of one of these exponents increases the defect strongly. | |
| Sep 6, 2013 at 18:09 | comment | added | S. Carnahan♦ | @Wolfgang I am unable to get all real eigenvalues in rank 7. Using a similar program, I have been unable to reduce the defect below 648 (which seems to be a broad local minimum), even after destroying the symmetry. Here is one of the minimal exponent collections: 0,59,75,99,108,132,52 | 0,27,101,107,111,124 | 0,85,107,109,104 | 0,88,105,104 | 0,40,73 | 0,56 | 0 | |
| Aug 12, 2013 at 0:06 | comment | added | S. Carnahan♦ | The "hull complex" is a tool in commutative algebra that is not particularly relevant to this problem. An explanation is in chapter 4 of Miller, Sturmfels Combinatorial Commutative algebra. I simply noticed that the construction of the hull complex uses a large parameter $T$ that is eventually thrown away. As far as my rank 6 answer goes, I decided to initialize my optimization by taking base 2 logs of the non-diagonal entries of your near-solution. Unsurprisingly, I didn't need to change much to get an actual solution. | |
| Aug 11, 2013 at 18:00 | comment | added | Wolfgang | Yes, I am fully with you. The "stretching" of these matrices (i.e. increasing $T$) doesn't seem to remove real eigenvalues, if T is big enough. I see that in rank 6, looking at the sequence 13,7,21,7,13 in your example, like me you did not yet succeed in finding a matrix such that all diagonals (in your notation) are "concave", let alone constant (i.e. 'triangular' Hankel matrices in my notation). But by the same "possibly irrational exuberance", I'd expect those to exist, too (like for $n\le 5$). When I have time, I'll play more. I don't understand though why you mention "complex hull"? | |
| Aug 11, 2013 at 11:50 | history | answered | S. Carnahan♦ | CC BY-SA 3.0 |