Timeline for Anti-bidiagonal matrix with main anti-diagonal {1,2,3,...} and first sub-anti-diagonal {-1,-2,-3,...} has eigenvalues lambda={1,-2,3,-4,...}
Current License: CC BY-SA 3.0
19 events
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| Oct 28, 2014 at 2:36 | comment | added | Richard Zhang | @AlexDegtyarev: Sorry that I didn't see your comment! To illustrate the difficulty, consider instead a tridiagonal matrix, whose characteristic polynomial has the recursion $P_n(x) = a P_{n-1}(x) - bc P_{n-2}(x)$. Suppose we knew $P_n(x_0) = 0$. Unfortunately, to show this in general would involve evaluating every $\{P_i(x_0)\}$, laboriously carrying the terms upwards, and seeing them magically cancel to zero at the final level. A naive attack via approach 1 in fact yields even more than 2 terms per level. | |
| Oct 28, 2014 at 1:10 | comment | added | Richard Zhang | @Suvrit: Yes I'm happy to fill you in on the details. Give me some time though to process everything ;-) | |
| Oct 28, 2014 at 1:01 | vote | accept | Richard Zhang | ||
| Oct 28, 2014 at 0:52 | comment | added | Richard Zhang | Many thanks to everyone for the helpful contributions. Ultimately, Darij and Suvrit's advice lead to a simple, short proof, which I will post once I've finished writing it. Essentially, Suvrit in a separate thread showed how a matrix like $B_n$ is related to a triangle matrix via a similarity transform, and this is a matrix whose columns are filled with the vector suggested in Darij's post. | |
| Oct 27, 2014 at 21:31 | comment | added | Suvrit | @RichardZhang: I am quite interested in such matrices, and it would be great if you could provide some concrete background / examples where such matrices arise. | |
| Oct 27, 2014 at 21:27 | review | Close votes | |||
| Oct 28, 2014 at 15:56 | |||||
| Oct 27, 2014 at 21:24 | answer | added | Suvrit | timeline score: 10 | |
| Oct 27, 2014 at 21:00 | comment | added | Suvrit | To me this seems like essentially a duplicate of: mathoverflow.net/questions/156090/… | |
| Oct 27, 2014 at 20:33 | comment | added | Alex Degtyarev | In your approach 1, why do you want to find the roots? Why not just plug in the suspected values? | |
| Oct 27, 2014 at 20:11 | history | edited | darij grinberg | CC BY-SA 3.0 |
added 1 character in body
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| Oct 27, 2014 at 19:52 | comment | added | darij grinberg | The right eigenvector for $\left(-1\right)^n n$ (the eigenvalue with largest absolute value) seems to be nice enough: $\left(\left(-1\right)^0 \binom{n-1}{0}, \left(-1\right)^1 \binom{n-1}{1}, \left(-1\right)^2 \binom{n-1}{2}, \cdots, \left(-1\right)^{n-1} \binom{n-1}{n-1}\right)$. The left eigenvector for the same eigenvalue seems to be $\left(\left(-1\right)^0 \binom{n}{0}, \left(-1\right)^1 \binom{n}{1}, \left(-1\right)^2 \binom{n}{2}, \cdots, \left(-1\right)^{n-1} \binom{n}{n-1}\right) ^T$. This suggests a recursive approach. | |
| Oct 27, 2014 at 19:50 | comment | added | Richard Zhang | Here are the eigenvectors plotted for the 6x6 case, as you can see, for low values of n, the functional view does not appear to be helpful. i.imgur.com/kstKaQ7.png | |
| Oct 27, 2014 at 19:46 | comment | added | Richard Zhang | They are polynomial-like. First is constant, second is linear, third is quadratic etc. More specifically, they can be viewed as sampled (or otherwise measured) versions of the continuous eigenfunctions. This is the same result you would get considering the continuous version (Approach 4). | |
| Oct 27, 2014 at 19:43 | comment | added | Dirk | Have you looked at the eigenvectors? | |
| Oct 27, 2014 at 19:35 | history | edited | Richard Zhang | CC BY-SA 3.0 |
wording & motivation
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| Oct 27, 2014 at 19:34 | comment | added | Richard Zhang | The question is: find a simple proof for Prop 1. Also, my apologies, by "one can show", I meant, "one can numerically verify". | |
| Oct 27, 2014 at 19:28 | comment | added | Noah Schweber | Is the question, "Find a simple proof that the eigenvalues are . . . "? Also, you say "one can show that this pattern persists." How does that proof go? | |
| Oct 27, 2014 at 19:27 | review | First posts | |||
| Oct 27, 2014 at 19:30 | |||||
| Oct 27, 2014 at 19:24 | history | asked | Richard Zhang | CC BY-SA 3.0 |
