TITLE: Homogeneous polynomials, mixed determinants, positive definiteness
QUESTION [10 upvotes]: 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.

REPLY [12 votes]: 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 $g_i$.