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How to prove that the determinant of a symmetric matrix with the main diagonal elements zero and all other elements positive is not zero (i.e., that the matrix is invertible)?

EDIT: OP indicates in a comment that the entries above the diagonal are to be distinct.

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    Just to make sure : do you mean a matrix of this form : $$ \begin{bmatrix} 0 & * & * & \dots \\ * & 0 & * & \dots \\ * & * & 0 & \dots \\ \vdots & \vdots & \vdots & \ddots \\ \end{bmatrix} $$ where the stars indicate positive entries?2012-05-12
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    Another very important thing : do your coefficients of your matrix lie in the real numbers? Or in some other field? I assume they do because you said "positive".2012-05-12
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    Maybe important remark : a computational experiment with all my stars replaced by ones suggest that the matrix with ones everywhere has determinant $n(-1)^n$ where $n$ is the size of the matrix. It also suggests that the determinant of such matrices will alternate signs when $n$ varies.2012-05-12
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    @tes: please edit the question to contain **all** relevant information. Right now one has to read a comment on one of the answers to find a rather significant hypothesis!2012-05-13

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