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Feb 22, 2023 at 20:27 comment added gondolf @Joseph Oh yes, thanks.
Feb 22, 2023 at 20:27 history edited gondolf CC BY-SA 4.0
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Feb 22, 2023 at 18:25 comment added Joseph Van Name Maybe you want $M^*M=N^*N$ but where there is no $\lambda\in\mathbb{C}^\times$ with $M=\lambda N$.
Feb 22, 2023 at 17:47 comment added Fabian Wirth $d=1$. Take $M=-N\neq 0$.
Feb 22, 2023 at 17:24 history asked gondolf CC BY-SA 4.0