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Find the determinant of the following matrix in the terms of $a_1,a_2,\cdots,a_n$ explicitly, $$ \begin{bmatrix} a_1 & a_2 & a_3 & \cdots & a_n\\ a_2 & a_3 & a_4 & \cdots & a_1\\ a_3 & a_4 & a_5 & \cdots & a_2\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ a_n & a_1 & a_2 & \cdots & a_{n-1}\\ \end{bmatrix} $$ When the determinant is zero?

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    $\begingroup$ en.wikipedia.org/wiki/Circulant_matrix $\endgroup$ – egreg Apr 26 '14 at 14:45
  • $\begingroup$ That was helpful ,thanks! $\endgroup$ – k1.M Apr 26 '14 at 14:52
  • $\begingroup$ @egreg: when I read the answer (and before reading your comment...) I suddenly thought of a community wiki, why not(?) $\endgroup$ – MattAllegro Apr 26 '14 at 21:55
  • $\begingroup$ See also: math.stackexchange.com/questions/81016/… $\endgroup$ – Martin Sleziak Feb 29 '16 at 13:12
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    $\begingroup$ We should remark that the matrix in question is not a circulant matrix, but an anti-circulant matrix. So, its determinant is $(-1)^{n(n-1)/2}$ (or $(-1)^{\lfloor n/2\rfloor}$) times the determinant of its circulant counterpart. $\endgroup$ – user1551 Feb 29 '16 at 15:02
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Call that matrix $A$ and notice that it is a permutation of a circulant matrix, $$ A = CP $$ Where $P$ is a permutation matrix with ones on the anti-diagonal, and zeros in all other positions. Then

$$ \det[A] = \det[CP] = \det[C]\det[P] $$ The determinant of the permutation part can be shown to depend on the size $n$. It can be written as $$ \det[P] = (-1)^{\left\lfloor\frac{n}{2}\right\rfloor} $$ Now $C$ is $$ \begin{bmatrix} a_{n} & a_{n-1} & a_{n-2} & \cdots & a_1\\ a_{1} & a_{n} & a_{n-1} & \cdots & a_2\\ a_{2} & a_{1} & a_n & \cdots & a_3\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ a_{n-1} & a_{n-2} & a_{n-3} & \cdots & a_{n}\\ \end{bmatrix} $$ $C$ is a circulant matrix. Define the associated polynomial $$ f(\omega) = a_n + \sum_{k=1}^{n-1} a_k\omega^k $$ Then using the product formula on the Wikipedia page for circulant matrices, $$ \det[C] = \prod_{j=0}^{n-1}f(\omega_j), $$ where $\omega_j=e^{\frac{2\pi i j}{n}}$ and $i=\sqrt{-1}$. Then the final formula is $$ \det[A] = (-1)^{\left\lfloor\frac{n}{2}\right\rfloor}\prod_{j=0}^{n-1}\left(a_n + \sum_{k=1}^{n-1} a_k\omega_j^k\right) $$

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  • $\begingroup$ $A$ is circular already and don't needed to use matrix $C$! $\endgroup$ – k1.M Apr 30 '14 at 11:38
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    $\begingroup$ $A$ is already a circulant? I don't believe so. The circulant marix form should at least have the same number on the main diagonal. $A$ does not have this property. Am I missing something? $\endgroup$ – rajb245 Apr 30 '14 at 20:33
  • $\begingroup$ @k1.M you might want to check out the definition on the Wikipedia page. $\endgroup$ – Vim Feb 29 '16 at 15:48
  • $\begingroup$ @k1.M as an alternative perspective, you can see the circulant matrices in $M_n(\Bbb K)$ as the subalgebra generated by the special circulant matrix $$\begin{bmatrix} 0 & 1 \\ I_{n-1} & 0\end{bmatrix}. $$ $\endgroup$ – Vim Feb 29 '16 at 15:52
  • $\begingroup$ Strange to see activity on this thread after so long, but I'm glad there are more eyes on the solution above. I never got any feedback from the original poster, positive or negative, other than his or her misunderstanding about the structure of circulant matrices. $\endgroup$ – rajb245 Mar 1 '16 at 2:07
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This answer is just the extension of the above rajb245's answer.

To answer more accurately also for the question "When the determinant is zero?" it's sufficient to quote Wikipedia.

Let the polynomial $ f(x) = a_n + a_1 x + \dots + a_{n-1} x^{n-1} $ be so called the ''associated polynomial'' of circulant matrix $C$. Matrix $C$ is defined as in the above rajb245's answer.

The rank of circulant matrix $ C $ is equal to $ n - d $, where $ d $ is the degree of a polynomial degree of $\gcd( f(x), x^n - 1) $.

So the determinant is equal to zero when $f(x)$ and $x^n-1$ have some common divisors.

This property can be used for constructing a circulant matrix which would be at the same time singular using fact that $x^n-1$ has real root $ \{1 \}$ for odd $n$ and $ \{ -1, +1 \}$ real roots for even $n$.

For example let $n=4$.
Then we have general form of associated polynomial for root $-1$.

$f(x)=(x+1)(b_3+b_1x+b_2x^2)= \\ b_3+(b_1+b_3)x+(b_2+b_1)x^2+b_2x^3 = \\ a_4 + a_1 x + a_2 x^2+ a_3 x^3 $

Take for example $f(x) = 6+7x+3x^2+2x^3$.

Indeed $\det{\begin{bmatrix} 6 & 2 & 3 & 7 \\ 7 & 6 & 2 & 3 \\ 3 & 7 & 6 & 2 \\ 2 & 3 & 7 & 6 \end{bmatrix}}=0$ (circulant matrix)

Also for corresponding anticirculant matrix (of the form as in the question)

$\det{\begin{bmatrix} 7 & 3 & 2 & 6 \\ 3 & 2 & 6 & 7 \\ 2 & 6 & 7 & 3 \\ 6 & 7 & 3 & 2 \end{bmatrix}}=0$.

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