Are there $n\times n$ real matrices $A\_{1}, \ldots, A\_{n}$ such that the $n$-homogeneous polynomial 
$$
f(x\_{1}, \ldots, x\_{n}) = \det(x\_{1} A\_{1}+\cdots +x\_{n} A\_{n})
$$ 
never vanishes on $\mathbb{R}^{n}\setminus\{0\}$?


I was listening to a seminar of a student, and a certain problem boils down to this linear algebraic question. I know that if $n$ is odd then the answer is negative; also if $n=2^{k}$ then the answer is positive. I do not quite see right now what happens for an arbitrary even $n$. The first interesting case is $n=6$. 


I believe this should be something well-known. 



Homogeneous polynomials, mixed determinants, positive definiteness
I doubt about powers of 2, it looks that the answer is $n=1,2,4,8$. 


Without loss of generality $A\_1=I$ (else replace $A\_i$ to $A\_iA\_1^{-1}$ for all $i$). Then for any $x\in \mathbb{S}^{n-1}$ the vectors $x=A\_1 x,A\_2x,\dots,A\_nx$ should be linearly independent (else $x$ belongs to a kernel of a certain linear combination of $A\_i$'s). Projecting $A\_2x,\dots,A\_nx$ onto the hyperplane $x^{\perp}$ we get for any $x$ an $(n-1)$-tuple of linearly independent vectors orthogonal to $x$, and they of course are continuous in $x$. That is, the sphere $\mathbb{S}^{n-1}$ is parallelizable, this is only the case for $n=1,2,4,8$. 


For these values of $n$, we take $n$-dimensional associative real division algebra (of real, complex, quaternionic or octavic numbers) generated by the lements $g\_1,\dots,g\_n$ and let $A\_i$ be the operator of right multiplication by $g\_i$. 


