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Let $n>3$ a natural number and let $a_{0},\,a_{1},\,\ldots,\,a_{n}$ be complex numbers such that $a_{0}+a_{1}+\ldots+a_{n}=0$. Is it possible to evaluate the following determinant: $$\left|\begin{array}{ccccc} a_{0} & a_{1} & a_{2} & \ldots & a_{n-1}\\ a_{1} & a_{2} & a_{3} & \ldots & a_{n}\\ a_{2} & a_{3} & a_{4} & \ldots & a_{0}\\ \vdots & \vdots & \vdots & \vdots & \vdots\\ a_{n-2} & a_{n-1} & a_{n} & \ldots & a_{n-4}\\ a_{n-1} & a_{n} & a_{0} & \ldots & a_{n-3} \end{array}\right|$$

(on every row and on every column is missing one element)?

thebalans
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  • What are your ideas so far? Have you tried computing the determinant for n=4, 5? – don-joe Feb 26 '18 at 12:53
  • The determinant is very close to a circulant determinant. (http://mathworld.wolfram.com/CirculantDeterminant.html). I tried to express the matrix as a product between a circulant matrix and another matrix which doesn't depend of $a_{0}, a_{1}, ...$ – thebalans Feb 26 '18 at 12:59
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    What you have is a special Hankel matrix. There is nothing special about the determinant if $a_0,a_1,\dots,a_{n-1}$ are arbitrary. – Somos Feb 26 '18 at 16:10
  • You're right. It is a special Hankel determinant. Unfortunately, I don't think that the Hankel determinant has a general evaluation. My problem is if (in this special case) it is possible to find an evaluation of it. – thebalans Feb 26 '18 at 16:30
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    In addition to what @don-joe stated in his comment, looking at your previous questions, you do not seem to show what you have attempted to solve the problems (assuming you have). Please include whether or not you have tried something yourself, so we can better help you. – Mr Pie Feb 27 '18 at 15:16
  • The determinant is like a circulant determinant without the last row and the last column. Trying for $n=4, 5...$ does not help too much. I believe that the $n-$root of 1 are involved, but I cannot find how. – thebalans Feb 27 '18 at 18:09

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