Timeline for The boundedness of $L_1$ norm $\|(I+A)^{-1}\|_1$ if both $\|A\|_1$ and $\|A^{-1}\|_1$ are bounded
Current License: CC BY-SA 4.0
8 events
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| Apr 10, 2023 at 12:30 | answer | added | Bazin | timeline score: 1 | |
| Apr 9, 2023 at 3:29 | history | edited | Golden Silence | CC BY-SA 4.0 |
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| Apr 9, 2023 at 2:15 | history | edited | Golden Silence | CC BY-SA 4.0 |
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| Apr 9, 2023 at 2:14 | comment | added | Golden Silence | Yes, for example, consider a unitary matrix where the first row is $N^{-1/2}(1,1,...,1)$. So it seems that the eigenvalue decomposition theorem cannot be utilized either. | |
| Apr 8, 2023 at 16:52 | comment | added | Christian Remling | @GeraldEdgar: The diagonal case is trivial ($\|(1+D)^{-1}\|_1\le 1$ when $D_{jj}\ge 0$), but the unitary matrix that diagonalizes $A$ could have large $1$ norm. | |
| Apr 8, 2023 at 11:01 | comment | added | Gerald Edgar | Can you do this in case $A$ is diagonal? Can the spectral theorem reduce the general case to the diagonal case? | |
| S Apr 8, 2023 at 8:48 | review | First questions | |||
| Apr 8, 2023 at 8:56 | |||||
| S Apr 8, 2023 at 8:48 | history | asked | Golden Silence | CC BY-SA 4.0 |