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Feb 11, 2023 at 23:47 history edited TanG CC BY-SA 4.0
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Feb 11, 2023 at 22:24 history edited TanG CC BY-SA 4.0
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Feb 11, 2023 at 20:39 vote accept TanG
Feb 11, 2023 at 20:31 history edited TanG CC BY-SA 4.0
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Feb 11, 2023 at 20:25 history edited TanG CC BY-SA 4.0
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Feb 10, 2023 at 23:31 answer added Saúl RM timeline score: 2
Feb 10, 2023 at 22:57 history edited TanG CC BY-SA 4.0
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Feb 10, 2023 at 22:57 history undeleted TanG
Jan 30, 2023 at 20:15 history deleted TanG via Vote
Jan 30, 2023 at 15:12 comment added TanG Yes, you are correct. My apology :) @GerryMyerson
Jan 30, 2023 at 15:08 comment added TanG Of course, @GerryMyerson you are correct. It was typo, I want to prove $n\leq 2k$, as Robert suggested, not $2n \leq k$. Sorry for the confusion.
Jan 30, 2023 at 15:07 history edited TanG CC BY-SA 4.0
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Jan 30, 2023 at 15:06 history edited TanG CC BY-SA 4.0
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Jan 30, 2023 at 0:35 comment added Robert Israel Using the canonical basis vectors and their negatives will give you $n = 2k$. Should the inequality be $n \le 2k$?
Jan 29, 2023 at 22:18 comment added Gerry Myerson $n=k=2$, $m_1=\mu_1=(1,0)$, $m_2=\mu_2=(-1,0)$ is a counterexample.
S Jan 29, 2023 at 22:04 review First questions
Jan 30, 2023 at 9:12
S Jan 29, 2023 at 22:04 history asked TanG CC BY-SA 4.0