Timeline for Smallest non-zero eigenvalue of a (0,1) matrix
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
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| Feb 14, 2014 at 2:55 | history | edited | Suvrit | CC BY-SA 3.0 |
reworded to make more precise..
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| Feb 14, 2014 at 2:52 | comment | added | Suvrit | Thanks Qiaochu! Yes, it is the algebraically the smallest eigenvalue, not in absolute value. Chasing that paper shows that characterizing the other eigenvalues (even for symmetric matrices) in an open problem. So I should add this caveat in the above answer and reword to correct. | |
| Feb 14, 2014 at 2:03 | comment | added | Qiaochu Yuan | $\lambda_{\text{min}}(A)$ is the most negative eigenvalue, not the smallest eigenvalue in absolute value. Note that this bound grows linearly in $n$ but we can clearly get eigenvalues of absolute value $1$ for all $n$ without much difficulty. | |
| Feb 14, 2014 at 1:32 | history | answered | Suvrit | CC BY-SA 3.0 |