A linear algebra problem about linear subspace of $GL_n$ 
Assume $A_1,\dots,A_n$ are fixed $n\times n$ real matrices and satisfy that for any nonzero vector $v ∈ \mathbb R^n$, the vectors $A_1v,\dots,A_nv$ form a basis for $\mathbb R^n$. Find the integers $n$ such that the matrices $A_1,\dots,A_n$ exist.
List examples of matrices $A_1,\dots, A_n$ for those $n$.

I found this problem in a book about linear algebra. I've solved the problem for $n=1,2,4$ and odd numbers before I posted the problem. At first I thought this problem is just about the matrices, but when observing the examples for $n=1,2,4$ I noticed that the problem has something to do with the product structure of $\mathbb R^n$ and this insight allows me to solve this problem.
 A: The answer is $n=1,2,4,8$.
The existence of $A_1,\dots,A_n$ is equivalent to the existence of a division composition algebra on $\mathbb R^n$.
This is because we can assume $A_1=I$ by substituting $v$ by $A_1^{-1}v$. Then fix a vector $v\neq 0$. Let $(v_1,v_2,\dots,v_n)$ denote  $(A_1v,\dots,A_nv)$ which is a basis of $\mathbb R^n$. Then we define $v_iw=A_iw$ for any $i$ and $w$, and using distribution law we can define $(a_1v_1+\dots+a_nv_n)w$ for all $(a_1,\dots,a_n)$ in $\mathbb R^n$. One can easily verify that this gives a division composition algebra of $\mathbb R^n$. By applying Hurwitz Theorem one knows that $1,2,4,8$ are the only answers.
A: I post this answer since I cannot post comments. I am not so sure about the $n=8$ case since the octonion algebra is not associative.
Furthermore, a composition algebra comes equipped with a nondegenerate quadratic form hence I am not sure we can apply the classification theorem for division algebras and exploit it for an existence result.
I think also that the assumption $A_1=I$ deserves clarification, for example in the case $n=2$ we can choose $A_1=\begin{pmatrix}1&0\\ 0&-1\end{pmatrix}$, $A_2=\begin{pmatrix}0&1\\ 1&0\end{pmatrix}$, even if it's true that we can multiply each $A_i$ by an invertible matrix $U$ and obtain a set with the same independence property, $I\notin\operatorname{span}\{A_1,A_2\}$.
