TITLE: Vector subspace of $M_n(\mathbb{R})$ with invertible matrices
QUESTION [10 upvotes]: I recall this claim that I have read in some book a long time ago, but now I do not remember and unfortunately I could not find anything on google about it. I was wondering if someone could help me with some reference about this. 

For $n>8$ there is no a $n$-dimensional vector subspace of $M_n(\mathbb{R})$ which all non zero elements are invertible matrix. 

I was also wondering if we can say something for $n \leq 8$. Thank you. 
Remark: I think this should be related to Hurwitz's Theorem (https://en.wikipedia.org/wiki/Hurwitz%27s_theorem_(composition_algebras)). For example, for $n=1,2,4,8$ these $n$-dimensional vector subspaces are the ones isomorphic to $\mathbb{R},\mathbb{C},\mathbb{H}$ (quaternions) and $\mathbb{O}$ (octonions) respectively. 
Remark 2: I think that the fact these matrices are real it is very important, but I don't know why.

REPLY [4 votes]: As to why it's important to work over $\mathbb{R}$, user1551 gives a first indication, but it goes in the "wrong" direction (it shows that there are even less subspaces with invertible matrices over $\mathbb{C}$).
If you take $\mathbb{Q}$, you can find $n$-dimensional subspaces of $M_n(\mathbb{Q})$ consisting of invertible matrices (except for $0$ of course) for arbitrarily large $n$. This is because just as over $\mathbb{R}$ you have the well-know Hamilton quaternions, over $\mathbb{Q}$ you have division algebras of all degrees (the degree is the squareroot of the dimension in this case, so quaternions are of degree $2$ and dimension $4$). To see that, you may for instance use the Brauer-Hasse-Noether theorem, but it is probably overkill (I just don't immediately see an elementary argument). 
If $D$ is such a division algebra, it embeds in $End_\mathbb{Q}(D)$ (where $D$ is seen as a vector space) by multiplication on the left, and the resulting matrices are invertible except for $0$.