Timeline for The influence of eigendecomposition on the periodicity of a (rank 2) Hermitian matrix (of functions)
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Aug 23, 2013 at 12:04 | comment | added | trienko | Of course $\boldsymbol{\Phi}$ is the $\phi$ only part of $\boldsymbol{R}$. | |
| Aug 23, 2013 at 12:03 | comment | added | trienko | Let $\mathcal{X}$ denote the set consisting of all square sub-matrices of $\boldsymbol{\Phi}$ (including $\boldsymbol{\Phi}$ itself) [obtained by deleting the same rows and columns <i.e. row 1 and column 1 and row 2 and column 2> of $\boldsymbol{\Phi}$]. Let $|\boldsymbol{A}|_{D} = \sum_{i=1}^n a_{i i+1}$ (where $n$ is the dimension of $\boldsymbol{A}$). Is a sufficient condition for $\boldsymbol{R}$ to be rank 2 that $\forall\boldsymbol{A}\in\mathcal{X}$ $|\boldsymbol{A}|_D = a_{1n}$, where $n$ is the dimension of $\boldsymbol{A}$?. | |
| Aug 23, 2013 at 10:16 | comment | added | trienko | I have checked your equation and as you said there is another restriction needed to be rank 2 (for 3 dimensional case). The equation $\phi_{13} - \phi_{12} - \phi_{23} = 0$. What is the general formula to make $\boldsymbol{R}$ rank 2 for any dimension? Is the original matrix positive semi-definite (so there is a largest eigenvalue)? Is the original proposition still true even though the matrix as stated above is not rank 2? | |
| Aug 23, 2013 at 7:47 | comment | added | trienko | Thank you very much for answering (I apologize for the wrong question), please (it would help me tremendously) give a numerical counter example. The weird thing is the matrices I am working with are all rank two. Meaning that there must be another property I am not seeing. | |
| Aug 22, 2013 at 15:06 | history | answered | Robert Israel | CC BY-SA 3.0 |