Timeline for Rayleight Ritz Ratio and smallest eigenvalue for a set of given matrices
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 9, 2012 at 9:57 | comment | added | dineshdileep | oh sorry, i just figured it out, $x$ should contain irrational entries in that case, isn't it? | |
| Dec 9, 2012 at 9:55 | comment | added | dineshdileep | @Denis I was reading more and more about numerical range. I found somethings hard to believe. Consider a single hermitian matrix $A_1$. So the image of the unit sphere will lie in the interval, $\lambda_{min}\leq x^HA_1x \leq \lambda_{max}$. So every irrational number in that interval will also come up for some non-zero $x$. But then how is that possible if $x$ and $A_1$ are not containing any irrational numbers. | |
| Dec 6, 2012 at 16:58 | comment | added | Denis Serre | @dineshdileep. If $A_1$ has a negative eigenvalue, ${\cal H}(M)$ has a point $w$ whose real part is negative. Likewise, if $A_2$ has a negative eigenvalue, ${\cal H}(M)$ has a point $z$ whose imaginary part is negative. From these facts, plus the convexity of ${\cal H}(M)$, you cannot conclude that it has a point with both coordinates negative. But of course, this will happen quite frequently. Example : if $M$ is normal, meaning that $A_1- and $A_2$ commute, then ${\cal H}(M)$ is just the convex hull of the eigenvalues of $M$: you see from the spectrum whether there is such a point. | |
| Dec 6, 2012 at 16:16 | comment | added | dineshdileep | @Denis Serre This gives me some more doubts. Since I took the matrices to be hermitian, it means both of them can have negative eigenvalues. In this case, does it mean, there will be points in that convex set, such that both the co-ordinates are negative?. Thanks to you, I am seeing the deep impact of the theorem. | |
| Dec 6, 2012 at 14:05 | comment | added | Denis Serre | @dineshdileep. Yes indeed. This is a bit surprising, because ${\cal H}(M)$ is the image of a non-convex set (the unit sphere) by a nonlinear map (the numerical map). But this is the contents of the TH Theorem. Actually, this is a deep result, with many interesting consequences and relations. For an extension of this result, see my recent paper with Th. Gallay in CPAM 65 (2012), pp 287-336. | |
| Dec 6, 2012 at 12:07 | comment | added | dineshdileep | @Denis Serre Actually I had one more doubt, the image is closed convex compact set, does it mean, every point in it has a corresponding vector in the unit sphere which gives that point? | |
| Dec 6, 2012 at 12:05 | comment | added | dineshdileep | @Felix Goldberg, I will take a look at it. | |
| Dec 6, 2012 at 12:05 | vote | accept | dineshdileep | ||
| Dec 6, 2012 at 12:05 | comment | added | dineshdileep | @Denis Serre Excellent answer!! Thanks for that one. | |
| Dec 5, 2012 at 11:55 | comment | added | Felix Goldberg | As a complement to Denis's approach, I think that joint numerical ranges could help to extend his alternative to $k$ matrices. See this paper, for example: sciencedirect.com/science/article/pii/S0024379503006797 | |
| Dec 5, 2012 at 10:42 | history | answered | Denis Serre | CC BY-SA 3.0 |