Circulant determinants

Question

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.)

It seems that it is anticirculant matrix. – Widawensen Jul 20 '17 at 16:10 — Widawensen, Jul 20 '17 at 16:10

score 18 · Accepted Answer · answered Apr 26 '14 at 16:49

18

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 $0$.

answered Apr 26 '14 at 16:49

Omran Kouba

29,112

2

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

score 6 · Answer 2 · answered Apr 26 '14 at 16:50

6

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

answered Apr 26 '14 at 16:50

Pipicito

2,800

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

Circulant determinants

2 Answers2

Linked