Timeline for Matrix inequality between a traceless matrix and identity
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 15, 2016 at 2:48 | vote | accept | Janus | ||
| Feb 14, 2016 at 12:19 | answer | added | Norbert Schuch | timeline score: 0 | |
| Feb 5, 2016 at 10:33 | review | Close votes | |||
| Feb 5, 2016 at 18:00 | |||||
| Feb 5, 2016 at 9:50 | comment | added | Fedor Petrov | In complex case this question now looks more difficult for me than I initially thought. It is not about matrices, you just have $n$ complex numbers $c_1,\dots,c_n$ such that $\sum c_i=0$, $\sum |c_i|=A$, $\sum |1+c_i|=B$ and you have to describe a locus $(A,B)\subset \mathbb{R}^2$. For example, if $n=2$, it is described by inequalities $\max(A,2)\leqslant B\leqslant \sqrt{A^2+4}$. | |
| Feb 5, 2016 at 9:07 | comment | added | Janus | @FedorPetrov Is there a good textbook to read for this? I don't have a solid background in matrix theory. I just ran across this problem while doing my research in quantum information theory. | |
| Feb 5, 2016 at 8:49 | comment | added | Fedor Petrov | At first, 'arbitrary field' and 'either $\mathbb{R}$ or $\mathbb{C}$' are different things. At second, it is a bad idea to denote the set of Hermitian matrices by $M_n(\mathbb{C})$. At third, for Hermitian matrix singular values are just absolute values of eigenvalues, thus we have to say something about $A=\sum |c_i|$ and $B=\sum |1+c_i|$, where $c_i$ are eigenvalues of $C$ and $\sum c_i=0$. It is an excersise to describe the locus of possible points $(A,B)$. | |
| Feb 5, 2016 at 8:18 | comment | added | Janus | @FedorPetrov Is there a method to check for a stricter bound on it? Thanks. | |
| Feb 5, 2016 at 8:17 | comment | added | Janus | @abx $\mathbb{F}$ is an arbitrary field, either $\mathbb{R}$ or $\mathbb{C}$. | |
| Feb 5, 2016 at 8:03 | comment | added | Fedor Petrov | Assuming that it is $\mathbb{C}$ (C and F are close on a keyboard), we may apply triangle inequality for the nuclear norm to conclude that they differ by at most $n$. | |
| Feb 5, 2016 at 7:56 | comment | added | abx | ... and what is $\Bbb{F}$? | |
| Feb 5, 2016 at 7:40 | comment | added | Janus | $C$ is a traceless Hermitian matrix, and $|C|$ is simply $\sqrt{C^{\dagger}C}$. Since $C$ is traceless, we know that tr$|C|$ is just the sum of the absolute values of the eigenvalues, which makes tr$|C|>0$. It will only be zero if $C=0$ and it is not. | |
| Feb 5, 2016 at 7:35 | comment | added | Fedor Petrov | What does $|C|$ mean? | |
| Feb 5, 2016 at 7:33 | review | First posts | |||
| Feb 5, 2016 at 7:41 | |||||
| Feb 5, 2016 at 7:29 | history | asked | Janus | CC BY-SA 3.0 |