# Circulant determinants

Suppose that $a_1,a_2,\ldots,a_n$ are $n$ distinct real numbers; is the following statement true?

There is a permutation of $a_1,a_2,\ldots,a_n$, namely $b_1,b_2,\ldots,b_n$, such that the determinant of the following matrix is nonzero: $$\begin{bmatrix} b_1&b_2&\cdots&b_n\\ b_2&b_3&\cdots&b_1\\ \vdots&\vdots&\ddots&\vdots\\ b_n&b_1&\cdots&b_{n-1}\\ \end{bmatrix}$$

(Such a matrix is called a circulant matrix.)

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This statement is not true, without supplementary conditions on the $a_i$'s. Indeed, suppose the $\sum_{k=1}^na_k=0$, whatever your permutation is the vector $[1,1,\ldots,1]^T$ is in the kernel of the circulant matrix of the $b_i$'s, and consequently, its determinant is $0$.

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Better answer than mine :P – Pipicito Apr 26 '14 at 16:51

Not in the general case. Take $a_1=-1$ and $a_2=1$. In any case the determinant will be $0$.

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Your answer is true but didn't covers general case! – k1.M Apr 26 '14 at 17:12