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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.

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  • $\begingroup$ How the answer can be negative for $n=1$? At first glance, a positive answer attained for all $n$ by taking matrices whose all $n^3$ entries are set-wise algebraicly independent. $\endgroup$
    – Max Alekseyev
    Commented Oct 5, 2016 at 1:34
  • $\begingroup$ Lets exclude the trivial case n=1. So what is your example when n=3? $\endgroup$
    – Paata Ivanishvili
    Commented Oct 5, 2016 at 1:38
  • $\begingroup$ For odd powers n>1 you get odd degree polynomials which have lots of zeros $\endgroup$
    – Paata Ivanishvili
    Commented Oct 5, 2016 at 1:41
  • $\begingroup$ @Max, the $x_i$ are not restricted to integral or rational or algebraic, so I don't see where the algebraic nature of the matrix entries comes into it. $\endgroup$
    – Gerry Myerson
    Commented Oct 5, 2016 at 1:41
  • $\begingroup$ May you provide an example for $n=2^k$? $\endgroup$
    – Ilya Bogdanov
    Commented Oct 5, 2016 at 8:05

1 Answer 1

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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 $g_i$.

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  • $\begingroup$ And if $A_1$ is degenerate, then $f(1, 0, \ldots, 0) = 0$... $\endgroup$
    – Ivan Izmestiev
    Commented Oct 5, 2016 at 8:35
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    $\begingroup$ Is there a simple analytic explanation of why $\mathbb{S}^{n-1}$ is not parallelizable when $n \neq 1,2,4,8$? For odd dimensions it follows from Milnor's simple proof of Hairy ball theorem. $\endgroup$
    – Paata Ivanishvili
    Commented Oct 5, 2016 at 14:46
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    $\begingroup$ I expect that no: there is neither no simple explanation, no analytical. I recommend lectures by V.V.Uspensky (in Russian) on the subject m.mathnet.ru/php/… $\endgroup$
    – Fedor Petrov
    Commented Oct 5, 2016 at 14:51

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