For which real $x$ is the matrix $ A_{ij} = x^{|i-j|}$ invertible?

Question

For which real $x$ is the $n\times n$ matrix with $ A_{ij} = x^{|i-j|}$ invertible?

Omitting the trivial cases $x=1,0$ I thought since it was real symmetric matrix I could diagonalize and look at its eigenvalues, but it quickly got messy and I couldn't come up with a closed solution

Answer 1

It seems if $A_n$ is such matrix of size $n \times n$ and $n \ge 1$, we have $$ \det A = \left(1-x^2\right)^{n-1}, $$ which is easy to prove using mathematical induction on Cramer's rule of expanding the determinant, expanding on the first column. The first submatrix is always the inductive hypothesis, and all the others except one can have an extra factor of $x$ factored out from the 1st row, which makes 1st and 2nd rows identical (and hence, determinant is 0).

Hence, for all $x \ne \pm 1$, the matrix $A_n$ is certainly invertible.

Answer 2

Let $A(n)$ be the $n \times n$ matrix. It looks to me like $\det(A(n)) = (1-x^2)^{n-1}$ for $n \ge 1$. Thus the only zeros are $\pm 1$.