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Sep 4, 2022 at 16:12 comment added Athere Thanks Christian, that's an excellent counterexample.
Sep 3, 2022 at 16:26 comment added Christian Remling The function $X\mapsto X^2$ is not operator monotone (search for these keywords perhaps for more information), so this fails even when $C=1$.
Sep 3, 2022 at 15:31 comment added Joseph Van Name Are you trying to show that $\forall A,B\exists C$ or $\forall A,B\forall C$? $\forall A,B\forall C$ can be easily disproven using computer calculations by testing random positive semidefinite matrices.
Sep 3, 2022 at 13:42 history asked Athere CC BY-SA 4.0