Timeline for Relating eigenvectors of two self-adjoints operators
Current License: CC BY-SA 3.0
3 events
| when toggle format | what | by | license | comment | |
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| Aug 8, 2012 at 9:21 | comment | added | Bramiozo | What kind off requirement would be in place for $\mathbf{R}_1$ and $\mathbf{R}_2$ for this to be true? | |
| Aug 7, 2012 at 15:28 | comment | added | Bramiozo | Thanks Fabien. I was wondering, suppose we reverse it and state that the set of eigenvectors $\mathbf{R}$ is the summation of two distinct parts, say $\mathbf{R}=\mathbf{R}_1+\mathbf{R}_2$ where each column represents an eigenvector. Now I want that $\mathbf{R}_1(i)\cdot\mathbf{R}_2(i)=0,\, \forall i$ where $i$ indicates a specific eigenvector $\mathbf{R}(i)$ and of course $\mathbf{R}(i)=\mathbf{R}_1(i)+\mathbf{R}_2(i)$. (Also suppose that the eigenvectors are normalised.) | |
| Jan 16, 2012 at 13:46 | history | answered | Fabien Besnard | CC BY-SA 3.0 |