# Determinant of circulant matrix

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?

• en.wikipedia.org/wiki/Circulant_matrix – egreg Apr 26 '14 at 14:45
• That was helpful ,thanks! – k1.M Apr 26 '14 at 14:52
• @egreg: when I read the answer (and before reading your comment...) I suddenly thought of a community wiki, why not(?) – MattAllegro Apr 26 '14 at 21:55
• – Martin Sleziak Feb 29 '16 at 13:12
• 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. – user1551 Feb 29 '16 at 15:02

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 $$\det[A] = (-1)^{\left\lfloor\frac{n}{2}\right\rfloor}\prod_{j=0}^{n-1}\left(a_n + \sum_{k=1}^{n-1} a_k\omega_j^k\right)$$

• $A$ is circular already and don't needed to use matrix $C$! – k1.M Apr 30 '14 at 11:38
• $A$ 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? – rajb245 Apr 30 '14 at 20:33
• @k1.M you might want to check out the definition on the Wikipedia page. – Vim Feb 29 '16 at 15:48
• @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 $$\begin{bmatrix} 0 & 1 \\ I_{n-1} & 0\end{bmatrix}.$$ – Vim Feb 29 '16 at 15:52
• 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. – rajb245 Mar 1 '16 at 2:07

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)

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