TITLE: Do there exist $n^2$ real coefficients such that all matrices in $\Bbb R^{n∗n}$ containing those coefficients are invertible?
QUESTION [5 upvotes]: Do there exist $n^2$ real coefficients such that all matrices in $\Bbb R^{n∗n}$ containing those coefficients (for any permutations) are invertible? Can we take these coefficients to be in $[1,2]$ ?

REPLY [9 votes]: You're trying to find $a = (a_{11}, \dots, a_{nn})\in \mathbb{R}^{n^2}$ such that the function
\begin{align*}
f(a) &= \det \begin{pmatrix}a_{11} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{nn}\end{pmatrix}
\end{align*}
has $f(ga)\not = 0$ for every permutation $g\in S_{n^2}$. Each $f(ga)$ is a polynomial in the $a_{ij}$ and thus is nonzero almost everywhere. Thus almost all $a$ satisfy your condition, and in particular you can choose a point in $[1, 2]^{n^2}$.