Let $A=(a_{ij})_{i,j=1}^n$ be a positive definite matrix with $a_{ij}>0$ for all $i, j$. Define the Hadamard inverse $A^{\circ -1}$ of $A$ as $(a_{ij}^{-1})_{i,j=1}^n$. Is it possible to decide whether $\det A^{\circ -1}$ is positive or negative?

  • $\begingroup$ The body asks for the determinant of the inverse, but the title asks for the determinant of the product of the matrix and the inverse. Please edit for consistency. $\endgroup$ – Gerry Myerson Feb 12 '17 at 12:12
  • 1
    $\begingroup$ @Gerry Myerson: Not any more. $\endgroup$ – Jan-Christoph Schlage-Puchta Feb 12 '17 at 12:29

I'm not sure what kind of answer you're looking for. There are $4\times4$ examples of positive definite symmetric matrices where the determinant of the Hadamard inverse is negative and examples where it is positive.

For the first type, take a $4\times4$ matrix that with all entries 1, and replace the main diagonal with 100's.

For the second type, start with the above and then replace the "anti-diagonal" (the entries (1,4), (2,3), (3,2), (4,1)) with 1/100.

Of course you can still decide if the determinant is positive or negative by computing it.

  • $\begingroup$ Is your second type positive definite? Doesn't it have a negative determinant? $\endgroup$ – Gerry Myerson Feb 12 '17 at 22:12
  • $\begingroup$ It still has the 100's on the main diagonal $\endgroup$ – Anthony Quas Feb 13 '17 at 1:16
  • $\begingroup$ OK, I missed that. $\endgroup$ – Gerry Myerson Feb 13 '17 at 1:35
  • $\begingroup$ but the second matrix is not positive definite. $\endgroup$ – Isha Garg Feb 13 '17 at 6:28
  • $\begingroup$ @IshaGarg: If the second matrix were required to be positive definite, it would automatically have a positive determinant and the question would be moot. $\endgroup$ – Anthony Quas Feb 14 '17 at 2:45

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