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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.

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

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You can think of rank of matrix.

If [mat A ] is to be invertible A has to to have full rank.

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  • $\begingroup$ It does look obvious that the additional row/column always increases the rank by $1$. But, how to prove it? $\endgroup$ – Axel Kemper May 12 at 10:09
  • $\begingroup$ I downvoted. This is not an answer to the question. $\endgroup$ – Giuseppe Negro May 12 at 10:26

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