Skip to main content

Timeline for An upper estimate for $|\det(A+B)|$

Current License: CC BY-SA 4.0

3 events
when toggle format what by license comment
Dec 29, 2018 at 0:24 vote accept Piotr Hajlasz
Dec 24, 2018 at 9:23 comment added Fedor Petrov we could use Hadamard inequality $|\det A|\leqslant \prod_{i=1}^n \|Ae_i\|\leqslant (\frac1n \sum \|Ae_i\|^2)^{n/2}=n^{-n/2} \|A\|_{HS}^n$
Dec 24, 2018 at 1:55 history answered Aleksei Kulikov CC BY-SA 4.0