Vector subspace of $M_n(\mathbb{R})$ with invertible matrices 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. 
 A: 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 $0$.
A: The maximal dimension of subspace of real invertible $n\times n$ matrices is given by the Hurwitz-Radon numbers $\rho(n)$, which is defined as follows: if $n=2^{4a+b}c$ where $0\le b\le3$ and $c$ is odd, then $\rho(n)=8a+2^b$. See

J. F. Adams (1962), Vector fields on spheres, Annals of Mathematics, 75(3): 603-632,
J. F. Adams, P. Lax and R. Phillips (1965), On matrices whose real linear combinations are non-singular, Proc. Amer. Math. Soc., 16:318-322,
J. F. Adams, P. Lax and R. Phillips (1966), Corrections to "On matrices whose real linear combinations are non-singular", Proc. Amer. Math. Soc., 17: 945-947.

The result of Adams (1962) essentially says that the maximal number of linearly independent vector fields on the $(n-1)$-dimensional sphere $S^{n-1}\subset\mathbb R^n$ is $\rho(n)-1$. The following presentation briefly explains the connection of these vector fields with invertible matrix subspace:

Rachel Quinlan, Special spaces of matrices, IMS Meeting 2013, NUI Maynooth.

Here are some obvious properties of the Hurwitz-Radon numbers: (a) $\rho(n)=1$ when $n$ is odd, (b) $\rho(n)\le n$ for every $n$, (c) $\rho(n)=n$ iff $n=1,2,4,8$. From (b) and (c), it follows that there is an $n$-dimensional subspace of real $n\times n$ invertible matrices if and only if $n=1,2,4,8$.
A proof of the weaker statement that the maximal dimension is bounded above by $n$ is given in

Zoran Z. Petrović (1999), On nonsingular matrices and Bott periodicity, Publications de l'Institut Mathématique 65(79).85: 97-102.

Clearly, the results of Adams et al. apply to real invertible matrices only. When the matrix space $M_n(\mathbb C)$, the maximal dimension of a subspace of invertible matrices is obviously $1$, as $A-\lambda B$ is singular when $\lambda$ is an eigenvalue of $AB^{-1}$.
