# Invertibility of the matrix whose elements are the cube of the distance of the indices.

I would like to prove, for any integer $$n>1$$, the invertibility of the $$n\times n$$ matrix $$A$$ whose elements are given by $$A_{ij}=|i-j|^3$$, where $$i$$ and $$j$$ are the indices.

To be clearer, for instance if $$n=5$$ I'm referring to the matrix $$A=\begin{bmatrix}&0&1&8&27&64\\&1&0&1&8&27\\&8&1&0&1&8\\&27&8&1&0&1\\&64&27&8&1&0\end{bmatrix}\,.$$

Such matrices should indeed be invertible (or at least it seems so looking at the determinant with Mathematica) for any $$n>1$$ and I do believe that there's should be some easy way to show it. I had some look on invertibility for Toeplitz or Hankel matrices, but I couldn't find any helps there for now.

• Perhaps you can show that the determinant is not divisible by some fixed integer; $n$ and $3$ are possibilities. – Aravind May 10 at 18:44

Actually I guess I've found a way to tackle the problem. Defining $$q$$ the matrix with elements $$q_{ij}=|i-j|$$, which is not hard to prove to be invertible, one has $$q^{-1}\,A\,q^{-1}=M\,,$$ where $$M$$ is a sparse matrix (nearly tridiagonal), which can be easily shown to be invertible.
Now I'm looking for a smart way to prove that $$M$$ has such a simple structure. It's not hard to do it by direct calculations but it's a bit tedious and I'm sure there must be some smarter way to do it.
• It does look obvious that the additional row/column always increases the rank by $1$. But, how to prove it? – Axel Kemper May 12 at 10:09