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 $ .



Problems concerning subspaces of $M_{n}(\mathbb{Q}) $
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 $n$.


