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Jun 8, 2021 at 19:20 vote accept Sky
May 17, 2021 at 17:42 comment added Vladimir Dotsenko Somehow, I am not sure that the tags are not misplaced.
May 17, 2021 at 17:32 history edited Sky CC BY-SA 4.0
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May 17, 2021 at 17:17 history edited Sky CC BY-SA 4.0
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May 17, 2021 at 9:36 comment added LSpice @Zero, do you want $B : (x, y, z) \mapsto (y, z, 2x)$? (I'm thinking of the obvious embedding of $\mathbb Q(\sqrt[3]2)$ in $\operatorname M_3(\mathbb Q)$, but maybe you have something else in mind!)
May 17, 2021 at 9:28 answer added Robert Bryant timeline score: 21
May 17, 2021 at 8:39 comment added user130903 This question does not have the same answer for $\mathbb Q$ as for $\mathbb R$. Let $A$ be the 3 x 3 diagonal matrix with entries (2,1,1) and let $B$ be the matrix of the linear map $(x,y,z)\mapsto (y,z,x)$, then $A$ and $B$ span such a space over $\mathbb Q$, since there is no third root of $2$ in $\mathbb Q$. However, $\rho(3)=1$.
May 17, 2021 at 8:25 comment added R.P. @LSpice Yes, informal suggestion. However you could define it presumably by intersecting the subspace with the unit hypersphere (in $\mathbb{R}^{n^2}$) and taking the distance between the intersections.
May 17, 2021 at 7:47 comment added LSpice This seems related to your earlier question mathoverflow.net/questions/392878/…. It's not a duplicate (at least I can't see how), but it might be better to figure out one question than to ask several closely related questions in succession.
May 17, 2021 at 7:45 comment added LSpice @RP_, how should one define closeness of subspaces? Or is it just an informal suggestion?
May 17, 2021 at 7:07 comment added R.P. The naive idea is that, since Q is dense in R, we can approximate a subspace N defined over R by a subspace N' defined over Q, and if N' is close enough to N then it should share the property of avoiding non-zero non-invertible matrices. Have you tried to make this work?
May 17, 2021 at 6:07 history asked Sky CC BY-SA 4.0