Real coefficients such that any matrix is invertible

Question

Let $n>1$. Does there exist $n^2$ real numbers $a_1, a_2, ..., a_{n^2}$ such that any $n \times n$ matrix using these coefficients is invertible?

I wanted to use the fact that a matrix is invertible iff its rank is $n$. My idea is to find numbers such that any column of $n$ of these numbers is not a linear combination of the other columns. So, I started looking at square roots of prime numbers. I was at first conjecturing that a sum of $n$ square roots of prime numbers cannot be itself a square root of prime numbers; It’s true for $n=2$:

If $p$ and $q$ are distinct prime numbers, let’s suppose by contradiction $\sqrt{p}+\sqrt{q}=\sqrt{k}$ with $k \in \mathbb{N}$. Squaring this identity gives $p+q+2\sqrt{pq}=k$ which is absurd since $\sqrt{pq}$ must be irrationnal.

For more primes, it gets more complicated since a similar proof for $n$ primes $p_1,...,p_n$ would lead to show that $\sum_{i<j} \sqrt{p_ip_j}$ cannot be a rationnal number, which I assume to be true but cannot prove it.

Moreover, proving the result only for sums cannot be sufficient since we’re looking for linear combinations, and there obviously exist a real $\lambda$ such that $\sqrt{p_1}=\lambda \sqrt{p_2}$, which makes impossible to do the study number by number as I was doing for now.

Could my idea lead anywhere, or is there an easier approach to this problem?

Answer 1

There are lots of them (almost any $n^2$ numbers works). Here is an explicit example. First, take $n^2$ natural numbers, such that all possible subset sums of these numbers are different. Say $\alpha_k = 2^{k-1}$, $k=1,\ldots, n^2$. Then take $a_k = \pi^{\alpha_k}$ ( instead of $\pi$, any transcendental number will do).