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Feb 10, 2023 at 0:48 comment added Dhruv Kohli Isn't $P$ the orthogonal projector onto $\mathcal{R}(A)^\perp$?
May 10, 2016 at 20:10 comment added Ulderique Demoitre I also found here sciencedirect.com/science/article/pii/S0024379509005230 this bound on the norms that is nice: $$ ||f(\Delta)|| \leq \mu \max({{||A||_2}^2,{||A_p||_2}^2})||\Delta ||_2 $$ where $\mu$ is a constant
May 9, 2016 at 16:44 vote accept Ulderique Demoitre
May 9, 2016 at 14:48 comment added Carlo Beenakker if $A$ is square and full rank, then $P=0$, but not so if $A$ is full row or column rank and non-square.
May 9, 2016 at 14:46 history edited Carlo Beenakker CC BY-SA 3.0
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May 9, 2016 at 13:59 comment added Ulderique Demoitre Thanks! If $A$ is full rank than $\Delta^\ast P=0$, correct? Do you have a reference for this?
May 9, 2016 at 10:24 history answered Carlo Beenakker CC BY-SA 3.0