What’s the sign of $\det\left(\sqrt{i^2+j^2}\right)_{1\le i,j\le n}$?

Suppose $A=(a_{ij})$ is a $n×n$ matrix by $a_{ij}=\sqrt{i^2+j^2}$. I have tried to check its sign by matlab. l find that the determinant is positive when n is odd and negative when n is even. How to prove it?

Answer

Recall that $e^{-|x|}=\int_0^\infty e^{-sx^2}\,d\mu(s)$ for some non-negative probability measure $\mu$ on $[0,+\infty)$ (Bernstein, totally monotone functions, etc.). Thus $e^{-t|x|}=\int_0^\infty e^{-st^2x^2}\,d\mu(s)$. for $t>0$. In particular, since $(e^{-st^2(i^2+j^2)})_{i,j}$ are rank one non-negative matrices, their mixture $(e^{-t\sqrt{i^2+j^2}})_{i,j}$ is a non-negative matrix for all $t>0$. Hence, considering the first two terms in the Taylor expansion as $t\to 0+$, we conclude that $-A$ is non-negative definite on the $n-1$-dimensional subspace $\sum_i x_i=0$, so the signature of $A$ is $1,n-1$. To finish, we just need to exclude the zero eigenvalue, i.e., to show that the columns are linearly independent, which I leave to someone else (i.e., up to this point we have shown that the determinant has the conjectured sign or is $0$).

Attribution
Source : Link , Question Author : Jimmy , Answer Author : fedja

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