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Consider an $n$ by $n$ whose diagonal has all zero entries, everything over the diagonal is -1 and under the diagonal is 1. Is there a neat way to compute the determinant? Looks like it is either 0 or 1 depending whether $n$ is even or odd and could make it into an inductive proof but maybe there is a better way?

Robert Barg
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  • @ancientmathematician, no, how would it? – Robert Barg Jan 21 '24 at 15:22
  • example $$ W = \left( \begin{array}{rr|rrrr} 0 & 1 & 1 & 1 & 1 & 1 \ -1 & 0 & 1 & 1 & 1 & 1 \ \hline -1 & -1 & 0 & 1 & 1 & 1 \ -1 & -1 & -1 & 0 & 1 & 1 \ -1 & -1 & -1 & -1 & 0 & 1 \ -1 & -1 & -1 & -1 & -1 & 0 \ \end{array} \right) $$ – Will Jagy Jan 21 '24 at 15:40
  • $$ R = \left( \begin{array}{rr|rrrr} 1 & 0 & 1 & 1 & 1 & 1 \ 0 & 1 & -1 & -1 & -1 & -1 \ \hline 0 & 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 0 & 0 & 1 \ \end{array} \right) $$ – Will Jagy Jan 21 '24 at 15:40
  • $$ R^T W R = \left( \begin{array}{rr|rrrr} 0 & 1 & 0 & 0 & 0 & 0 \ -1 & 0 & 0 & 0 & 0 & 0 \ \hline 0 & 0 & 0 & 1 & 1 & 1 \ 0 & 0 & -1 & 0 & 1 & 1 \ 0 & 0 & -1 & -1 & 0 & 1 \ 0 & 0 & -1 & -1 & -1 & 0 \ \end{array} \right) $$ – Will Jagy Jan 21 '24 at 15:41
  • see Newman, Integral Matrices, especially the skew normal form, pages 56-59 in Chapter IV, Congruence. The comments above illustrate induction, increasing $n$ by $2 ; ; ; $ – Will Jagy Jan 21 '24 at 15:45
  • @RobertBarg I was wrong in suggesting it was a duplicate. I have voted to re-open. – ancient mathematician Jan 22 '24 at 07:39

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