Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.)

share|improve this question

2 Answers 2

up vote 16 down vote accepted

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 $0$.

share|improve this answer
Better answer than mine :P –  Pipicito Apr 26 at 16:51

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

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.