Timeline for Add a multiple of $I$ to a matrix to minimize its operator norm
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Feb 21, 2016 at 13:08 | answer | added | Suvrit | timeline score: 4 | |
| Feb 21, 2016 at 11:50 | history | edited | Federico Poloni | CC BY-SA 3.0 |
changed notation to give $s$ a different name
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| Feb 21, 2016 at 11:48 | comment | added | Federico Poloni | @PietroMajer Yes, if $A$ is normal then that is the solution (via a diagonalization argument). This also shows that in general the Frobenius-norm minimizer is different from the operator-norm minimizer: just take a $3\times 3$ diagonal matrix whose diagonal entries form a triangle with centroid different from its circumcenter. | |
| Feb 21, 2016 at 11:09 | comment | added | Lior Silberman | Conjugation by a unitary matrix doesn't affect the operator norm either, so you may assume $A$ is 70034-triangular. If $A$ is diagonal it's operator norm is the largest entry, so indeed in the normal case take $s$ to be the circumcentre of the spectrum. This may hold in general. | |
| Feb 21, 2016 at 9:31 | comment | added | Pietro Majer | in the case of A normal, I guess one gets s = the center of the minimum disk containing spec(A), right? | |
| Feb 21, 2016 at 8:31 | history | asked | Federico Poloni | CC BY-SA 3.0 |