11
$\begingroup$

Let $M_{n}(\mathbb{Q}) $ denote the $n$ times $n$ matrices over the rational number field. $N$ be a subspace of $M_{n}(\mathbb{Q}) $.Then if all the non-zero matrices in $N$ are invertible, what is the maximum the dimension of $N$ can be?

We already know that if we take $M_{n}(\mathbb{R}) $ instead of $M_{n}(\mathbb{Q}) $ then the answer is $ \rho(n) $. where $ \rho(n) $ is Radon Hurwitz number i.e if $ n = (2a + 1 ) 2^{c+4d} $ where $ 0 \leq c \leq 3 $ then $ \rho(n) = 2^{c} + 8d $ .

Improve this question
$\endgroup$
7
  • $\begingroup$ 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? $\endgroup$
    – R.P.
    Commented May 17, 2021 at 7:07
  • $\begingroup$ @RP_, how should one define closeness of subspaces? Or is it just an informal suggestion? $\endgroup$
    – LSpice
    Commented May 17, 2021 at 7:45
  • 1
    $\begingroup$ 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. $\endgroup$
    – LSpice
    Commented May 17, 2021 at 7:47
  • 5
    $\begingroup$ 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, $\rho(3)=1$. $\endgroup$
    – user130903
    Commented May 17, 2021 at 8:39
  • 1
    $\begingroup$ @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!) $\endgroup$
    – LSpice
    Commented May 17, 2021 at 9:36

1 Answer 1

Reset to default
21
$\begingroup$

Let's call this maximal dimension function $\rho_{\mathbb{Q}}:\mathbb{N}\to\mathbb{N}$, i.e., $\rho_{\mathbb{Q}}(n)$ is the largest possible dimension of a subspace $N\subset M_n(\mathbb{Q})$ such that all of the nonzero elements of $N$ are invertible.

Then $\rho_{\mathbb{Q}}(n)\ge n$, as the following construction shows: Let $p(x)\in\mathbb{Q}[x]$ be a polynomial of degree $n$ that is irreducible over $\mathbb{Q}$, and let $A\in M_n(\mathbb{Q})$ be a matrix whose characteristic polynomial is $p$. (Such $A$ are easily constructed.) Let $N\subset M_n(\mathbb{Q})$ be the $\mathbb{Q}$-subspace spanned by the powers of $A$. Then $N$ is an $n$-dimensional over $\mathbb{Q}$ and $N$ is a field isomorphic to $\mathbb{Q}[x]/\bigl(p(x)\bigr)$, so every non-zero element of $N$ is invertible.

Meanwhile, it's easy to see that $\rho_{\mathbb{Q}}(2)\le 2$, since any $3$-dimensional subspace of $M_2(\mathbb{Q})$ contains a nonzero element with vanishing determinant. Thus, $\rho_{\mathbb{Q}}(2) = 2$. Moreover, as Fedor points out in his comment below, this observation extends to all $n$ because, if $N\subset M_n(\mathbb{Q})$ had dimension greater than $n$ over $\mathbb{Q}$, then $N$ would have to intersect the codimension $n$ subspace of $M_n(\mathbb{Q})$ consisting of those matrices with first column equal to $0$. Thus, $\rho_\mathbb{Q}(n)=n$ for all $n$.

Improve this answer
$\endgroup$
15
  • 5
    $\begingroup$ but any space of dimension $n+1$ contains a non-zero matrix with zero first column, right? $\endgroup$
    – Fedor Petrov
    Commented May 17, 2021 at 10:29
  • 1
    $\begingroup$ @FedorPetrov: Yes, indeed, which implies that $\rho_{\mathbb{Q}}(n)\le n$. Thanks. $\endgroup$
    – Robert Bryant
    Commented May 17, 2021 at 10:39
  • 4
    $\begingroup$ Dear Sugata, this is a new question, please ask it as such. $\endgroup$
    – Fedor Petrov
    Commented May 17, 2021 at 15:32
  • 5
    $\begingroup$ @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. $\endgroup$
    – R.P.
    Commented May 17, 2021 at 17:25
  • 1
    $\begingroup$ okaay I have cleaned my edited part $\endgroup$
    – Sky
    Commented May 17, 2021 at 17:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .