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In Dimers and Amoebae by Kenyon, Okounkov, and Sheffield (2003), they say that it is easy to see that for matrices of the form

$$ \left( \begin{array}{ccccc} a_1 & 0 & 0 & 0 & b_n \\ b_1 & a_2 & 0 & 0 & 0 \\ 0 & b_2 & a_3 & 0 & 0 \\ 0 & 0 & b_3 & \ddots & 0 \\ 0 & 0 & 0 & \ddots & a_n \end{array} \right), $$

where $\{a_1,\dots, a_n\}$ and $\{b_1,\dots, b_n\}$ are nonnegative numbers, all odd sized minors are nonnegative. Does anybody have a way to nicely show this?

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    $\begingroup$ A couple of ideas. If we can show this is true for positive entries $a_i$ and $b_i$, then by continuity of the determinant (as a polynomial in the entries) it is true for nonnegative $a_i,b_i$. Also, we can normalize away either the $a_i$ or the $b_i$ in the case that these are positive by factoring one of them out of each column. $\endgroup$ – hardmath Jul 13 '15 at 12:29
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    $\begingroup$ A couple of other ideas: If $b_n=0$, the matrix is the weighted path matrix of a planar network (which this comment is too small to contain), hence totally nonnegative. So the only minors that might be negative are the ones containing $b_n$. Maybe it's possible to argue directly, since there are so many zero entries, that such a minor is either the whole matrix ($\prod(a_k b_k)$ if $n$ is odd), or else just contains a single term (i.e., a contribution from only one permutation, of positive sign). $\endgroup$ – Hans Lundmark Jul 13 '15 at 17:54
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    $\begingroup$ Sorry, what I meant was of course that the whole determinant is $\prod a_k + (-1)^{n-1} \prod b_k$. $\endgroup$ – Hans Lundmark Jul 13 '15 at 18:37

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