Timeline for The eigenvalues of the sum of two nilpotent matrices
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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| Jan 5, 2011 at 18:28 | comment | added | David E Speyer | Oh, I see. You're still not going to get anywhere. Conditions on how $F$ behaves on the unit circle are only conditions on $F$ modulo $x^2+y^2-1$. So, at this point, you have imposed: F = 1 mod x^2+y^2, F=(something related to Chebyshev polynomials) mod x^2+y^2-1, and F hyperbolic. That still won't force F to be a function of x^2+y^2. There should be a counterexample in degree $5$: Take one of your polynomials and perturb it by $\epsilon x(x^2+y^2)(x^2+y^2-1)$. For $\epsilon$ small, you haven't broken hyperbolicity, and you haven't changed its values on the unit circle, or on $x \pm i y=0$. | |
| Jan 5, 2011 at 17:31 | comment | added | Eslam | the condition I added is not only that it is even but that only component which are multiples of $\frac{n}{2}$ are non-vanishing. | |
| Jan 5, 2011 at 17:20 | vote | accept | Eslam | ||
| Jan 5, 2011 at 17:31 | |||||
| Jan 5, 2011 at 17:09 | comment | added | David E Speyer | Oh and, of course, you can easily embed Denis's example into a $4 \times 4$ example by padding with $0$'s. | |
| Jan 5, 2011 at 17:07 | comment | added | David E Speyer | Continuing to use Denis's very nice coordinates, we have $X=\cos \theta$ and $Y = \sin \theta$. Your added condition is that $F(\cos \theta, \sin \theta)$ has all Fourier coefficients even or, in other words, that it is an even function of $\theta$. That means there must be no odd powers of $y$. Denis's polynomial has that property. | |
| Jan 5, 2011 at 17:04 | history | edited | Denis Serre | CC BY-SA 2.5 |
added 585 characters in body
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| Jan 5, 2011 at 16:51 | comment | added | Eslam | Thanks for the reply. Seemingly all the details of my particular problem has to do with the fact that my eigenvalues come out independent on q for all even dimensions I tried so far. For example one more thing I deduced is that in my problem (whether or not the matrices are nilpotent) the eigenvalues can be written as $\epsilon_k(q) = \sum_{l=0}^{\infty} a_{kl} cos(\frac{n}{2}lq)$ where n is the matrix dimension (which is always even) | |
| Jan 5, 2011 at 16:42 | comment | added | David E Speyer | To connect our ideas together, the point here is that there are plenty of polynomials $F(x,y)$ such that $F$ is hyperbolic, and $F \equiv 1 \mod x^2+y^2$, but $F$ is not a function of $x^2+y^2$. For example, your matrix gives $1-3(x^2+y^2)+2x (x^2+y^2)$ | |
| Jan 5, 2011 at 16:32 | history | answered | Denis Serre | CC BY-SA 2.5 |