Timeline for Maximum spectral norm of matrices with given anti-Hermitian part and Hermitian part's spectrum
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Oct 2, 2017 at 19:02 | comment | added | Christian Remling | In other words, the only statement that one can make for general $A,H$ is that the max will be between $(\|A\|^2+\|H\|^2)^{1/2}$ and $\|A\|+\|H\|$. | |
| Oct 1, 2017 at 21:41 | comment | added | Christian Remling | I'm not sure one can give a very general answer, it'll depend on what $A,H$ are. If both matrices have one eigenvalue that is much larger than all the others, then what Yemon suggests feels about right. If they have eigenvalues of comparable (or equal) size, then one has more options and can do better. | |
| Oct 1, 2017 at 21:39 | comment | added | Christian Remling | @YemonChoi: Actually your bound is not optimal in general, sometimes $\|A\|+\|H\|$ is possible. | |
| Oct 1, 2017 at 21:04 | comment | added | Yemon Choi | francesco999: sorry I wrote my comments in a rush, hence the typo and the lack of explanation. @ChristianRemling: I can't see how to get the upper bound you mention, since $A$ is skew Hermitian and $H$ is Hermitian | |
| Oct 1, 2017 at 20:11 | comment | added | francesco999 | @YemonChoi I am going to guess $U^H U^*$ actually stands for $UHU^*$. Notice that $(x, M_U^* M_U x)=(x,U^*H^2U x)+(x,A^2x)+(x,[U^*HU,A]x)$ so choosing $x$ to be the (normalized) eigenvector of maximum eigenvalue (in modulus) of $A$ and $Ux$ the same for $H$ will maximize the first two terms, but not necessarily the third? WLOG we may assume at least that both $A$ and $H$ are diagonal matrices. | |
| Oct 1, 2017 at 19:57 | comment | added | francesco999 | Also, it would be great if you could spell out maybe in more detail how you get to $(||H||^2+||A||^2)^{1/2}$ | |
| Oct 1, 2017 at 19:55 | comment | added | francesco999 | I am not familiar with the notation $U^H$. What does it mean? | |
| Oct 1, 2017 at 19:48 | history | edited | francesco999 | CC BY-SA 3.0 |
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| Oct 1, 2017 at 16:07 | comment | added | Yemon Choi | In which case I think you get $(\Vert H\Vert^2+\Vert A\Vert^2)^{1/2}$ | |
| Oct 1, 2017 at 16:06 | comment | added | Yemon Choi | Can't you just choose $U$ to "rotate" things such that an eigenvector of $A$ corresponding to its largest (in modulus) eigenvalue also becomes an eignvector of$ U^HU^*$ corresponding to its largest (in modulus) eigenvalue? | |
| Oct 1, 2017 at 16:05 | review | First posts | |||
| Oct 1, 2017 at 16:11 | |||||
| Oct 1, 2017 at 16:03 | history | asked | francesco999 | CC BY-SA 3.0 |