TITLE: Determinant of "quasi-circulant" matrices
QUESTION [6 upvotes]: Several weeks ago I asked a question regarding the determinant of a $5 \times 5$ matrix: Determinant of a matrix when its diagonal elements have a certain property
Here I would like to consider a more general problem:the non-diagonal element will still share a circulant fashion, whereas some of them can be $0$ instead of all being $-1$. To be more specific, it follows a pattern that in each row, the first $i^{th}$ elements that follow the diagonal element are $-1$ while the rest are $0$.
For instance, in a $5\times5$ case, things can be
$A=\begin{pmatrix}
l_1&-1 &-1 &-1&0 \\
0&l_2 &-1 & -1 & -1\\
-1 & 0 &l_3 & -1 &-1\\
-1 & -1 & 0 & l_4 & -1\\
-1 & -1 &-1 & 0 &l_5
\end{pmatrix}$
or
$A=\begin{pmatrix}
l_1&-1 &-1 &0 &0 \\
0&l_2 & -1 & -1 & 0\\
0 & 0 &l_3 & -1 &-1\\
-1 & 0 & 0 & l_4 & -1\\
-1 & -1 & 0 & 0 &l_5
\end{pmatrix}$
Again, I would like to know if there is any delicate way to show that det$(A)<0$ when one of the $l_i$ equals to $0$ whereas others are positive.
Some of my thoughts:
Consider $l_1=l_2=...=l_n=l$($n$ is the dimension of the matrix). In this case, $A$ is a circulant matrix, and det($A$) can be demonstrated as a polynomial of $l$. It suffices to show that $l^n$ is the only term with positive coefficient, yet for me this is something too tedious to show.(with all those roots of unity to deal with)
I have also tried the matrix determinant lemma, which is helpful when all the non-diagonal elements are $-1$ but not that helpful when some of the non-diagonal elements are $0$.
Any hints or ideas will be greatly appreciated. Thanks!
Sorry for not making this clear in the first place, but this is just a conjecture (though I believe this is correct). A counterexample will also be greatly appreciated!

REPLY [2 votes]: Here is a counter example:
$$
\left[\begin{matrix}0 & 0 & -1.0 & 0 & -1.0 & -1.0\\-1.0 & 0.391134657778849 & 0 & -1.0 & 0 & -1.0\\-1.0 & -1.0 & 0.391134657778849 & 0 & -1.0 & 0\\0 & -1.0 & -1.0 & 0.391134657778849 & 0 & -1.0\\-1.0 & 0 & -1.0 & -1.0 & 0.391134657778849 & 0\\0 & -1.0 & 0 & -1.0 & -1.0 & 0.391134657778849\end{matrix}\right]
$$
It has a determinant of 10.609672399765413.
Here is another one (determinant of 0.99):
$$
\left[\begin{matrix}0 & 0 & -1.0 & 0\\0 & 0.10 & 0 & -1.0\\-1.0 & 0 & 0.10 & 0\\0 & -1.0 & 0 & 0.10\end{matrix}\right]
$$