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Jun 8, 2021 at 19:20 vote accept Sky
Jun 8, 2021 at 19:15 comment added Robert Bryant @SugataMandal: I looked and found this link: mathoverflow.net/help/someone-answers. Apparently, you (and only you) should see a greyed out check mark near the vote tally for an answer to your question, and what you do is just click that to turn it green, which constitutes 'accepting' that answer.
Jun 8, 2021 at 19:10 comment added Robert Bryant @SugataMandal: That's interesting. Just voting it up does not constitute 'accepting' an answer. My answer is certainly not officially accepted, because there's no 'green check mark' underneath the vote tally, the way there is when the question owner accepts an answer. (For example, look at mathoverflow.net/questions/394526/…) You, as the question owner, must have some special link to click somewhere that does this, but, since I have never asked a question, I don't know what it would look like.
Jun 8, 2021 at 18:39 comment added Sky @RobertBryant actually I dont know the procedure , but I think I already accepted your answer , I could not find any option for acceptance other than voting up .
Jun 8, 2021 at 15:30 comment added Robert Bryant @SugataMandal: If there is something that you find unsatisfactory about my answer to this question, could you say what it is? Otherwise, you should accept it, so that people won't be wondering what's wrong with it. (It's cleaner than leaving things hanging.) Also, you might consider re-titling the question so that it reflects what it's really about. The current title is kind of vague.
May 17, 2021 at 18:14 comment added Sky Yes thanks for suggetion . I have already asked my question as a fresh question.
May 17, 2021 at 17:41 comment added Vladimir Dotsenko @SugataMandal this is an exhaustive answer to your question. You should thus accept it, and ask a new question if you wish to get an answer to a different question.
May 17, 2021 at 17:34 comment added Sky okaay I have cleaned my edited part
May 17, 2021 at 17:25 comment added R.P. @SugataMandal If you edit your question after it was answered, things get confusing very fast. It is no longer clear that Robert Bryant's answer answers the whole of your original question, and it is also not clear which part of your question he is answering, because the numbering was only introduced in the edit. And then there can be multiple partial answers of which you can only accept one. So much better to keep things clean and ask a new question.
May 17, 2021 at 17:19 comment added Sky I have edited this question @FedorPetrov as they are related .
May 17, 2021 at 15:32 comment added Fedor Petrov Dear Sugata, this is a new question, please ask it as such.
May 17, 2021 at 15:30 comment added Sky Yes @FedorPetrov ,@Robert the way of proof is correct but now I am interested in the case if all the ellements in $N$ is skew symmetric and invertible then what will be the maximal dimension of $N$. You can consider $ n$ is even.. Note that if we take $\mathbb{R} $ instead of $\mathbb{Q} $ then for $ n=4 $ and $n= 8 $ the answer is 3 and 7 respectively, for $ n= 6 $ there does not exist any $N $ subspace of $M_{6}(\mathbb{R}) $ such that dim$N= 5$.
May 17, 2021 at 10:42 history edited Robert Bryant CC BY-SA 4.0
Incorporated Fedor Petrov's comment to complete the solution.
May 17, 2021 at 10:39 comment added Robert Bryant @FedorPetrov: Yes, indeed, which implies that $\rho_{\mathbb{Q}}(n)\le n$. Thanks.
May 17, 2021 at 10:29 comment added Fedor Petrov but any space of dimension $n+1$ contains a non-zero matrix with zero first column, right?
May 17, 2021 at 10:07 history edited Robert Bryant CC BY-SA 4.0
Changed notation for clarity.
May 17, 2021 at 10:06 comment added Robert Bryant @JensReinhold: OK. I'll call it $\rho_{\mathbb{Q}}$.
May 17, 2021 at 10:02 comment added Jens Reinhold Using $\rho$ in this post is misleading as it is used differently in the question. The function over $\mathbb Q$ has no name there.
May 17, 2021 at 9:56 history edited Robert Bryant CC BY-SA 4.0
Added a comment settling the case n=2
May 17, 2021 at 9:28 history answered Robert Bryant CC BY-SA 4.0