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DoAre there exist $n\times n$ real matrices $A_{1}, \ldots, A_{n}$ such that the $n$-homogeneous polynomial $$ f(x_{1}, \ldots, x_{n}) = \det(x_{1} A_{1}+\cdots +x_{n} A_{n}) $$ is never zeronever 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, andnegative; also if $n=2^{k}$ then the answer is positive. I do not quite see right now whatwhat happens for an arbitrary even $n$. The first interesting case is $n=6$.

I believe this should be something well known-known.

Do there exist $n\times n$ real matrices $A_{1}, \ldots, A_{n}$ such that $n$-homogeneous polynomial $$ f(x_{1}, \ldots, x_{n}) = \det(x_{1} A_{1}+\cdots +x_{n} A_{n}) $$ is never zero on $\mathbb{R}^{n}\setminus\{0\}$?

I was listening to a seminar of a student, and a problem boils down to this linear algebraic question. I know that if $n$ is odd the answer is negative, and if $n=2^{k}$ 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.

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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Paata Ivanishvili
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Homogeneous polynomials, mixed determinants, positive definiteness

Do there exist $n\times n$ real matrices $A_{1}, \ldots, A_{n}$ such that $n$-homogeneous polynomial $$ f(x_{1}, \ldots, x_{n}) = \det(x_{1} A_{1}+\cdots +x_{n} A_{n}) $$ is never zero on $\mathbb{R}^{n}\setminus\{0\}$?

I was listening to a seminar of a student, and a problem boils down to this linear algebraic question. I know that if $n$ is odd the answer is negative, and if $n=2^{k}$ 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.