# Determinant of a General Circulant Matrix ## Problem 374

Let $A=\begin{bmatrix} a_0 & a_1 & \dots & a_{n-2} &a_{n-1} \\ a_{n-1} & a_0 & \dots & a_{n-3} & a_{n-2} \\ a_{n-2} & a_{n-1} & \dots & a_{n-4} & a_{n-3} \\ \vdots & \vdots & \dots & \vdots & \vdots \\ a_{2} & a_3 & \dots & a_{0} & a_{1}\\ a_{1} & a_2 & \dots & a_{n-1} & a_{0} \end{bmatrix}$ be a complex $n \times n$ matrix.
Such a matrix is called circulant matrix.
Then prove that the determinant of the circulant matrix $A$ is given by
$\det(A)=\prod_{k=0}^{n-1}(a_0+a_1\zeta^k+a_2 \zeta^{2k}+\cdots+a_{n-1}\zeta^{k(n-1)}),$ where $\zeta=e^{2 \pi i/n}$ is a primitive $n$-th root of unity. Add to solve later

## Proof.

Let $\omega$ be any $n$-th root of unity.
Consider the vector
$\mathbf{v}=\begin{bmatrix} 1 \\ \omega \\ \omega^2 \\ \vdots \\ \omega^{n-1} \end{bmatrix}.$ We show that the vector $\mathbf{v}$ is an eigenvector of $A$.
We compute
\begin{align*}
A\mathbf{v}=\
\begin{bmatrix}
a_0 & a_1 & \dots & a_{n-2} &a_{n-1} \\
a_{n-1} & a_0 & \dots & a_{n-3} & a_{n-2} \\
a_{n-2} & a_{n-1} & \dots & a_{n-4} & a_{n-3} \\
\vdots & \vdots & \dots & \vdots & \vdots \\
a_{2} & a_3 & \dots & a_{0} & a_{1}\\
a_{1} & a_2 & \dots & a_{n-1} & a_{0}
\end{bmatrix}
\begin{bmatrix}
1 \\
\omega \\
\omega^2 \\
\vdots \\
\omega^{n-1}\\
\end{bmatrix}.
\end{align*}

The first entry of the vector $A\mathbf{v}$ is
$a_0+a_1\omega+a_2\omega^2+\cdots a_{n-2}\omega^{n-2}+a_{n-1}\omega^{n-1}=:\lambda.$ We define $\lambda$ to be this number.

The second entry is
\begin{align*}
&a_{n-1}+a_0\omega+\cdots+a_{n-3}\omega^{n-2}+a_{n-2}\omega^{n-1}\\
&=(a_{n-1}\omega^{n-1}+a_0+\cdots+a_{n-3}\omega^{n-3}+a_{n-2}\omega^{n-2})\omega \\
&=(a_0+\cdots+a_{n-3}\omega^{n-3}+a_{n-2}\omega^{n-2}+a_{n-1}\omega^{n-1})\omega \\
&=\lambda \omega
\end{align*}

Similarly the $i$-th entry of the vector $A\mathbf{v}$ is
\begin{align*}
&a_{n-i+1}+a_{n-i+2}\omega +\cdots+ a_{n-i}\omega^{n-1}\\
&= (a_{n-i+1}\omega^{n-i+1}+a_{n-i+2}\omega^{n-i+2} +\cdots+ a_{n-i}\omega^{n-i})\omega^{i-1}\\
&=\lambda \omega^{i-1} .
\end{align*}
Therefore we obtain
$A\mathbf{v}=\begin{bmatrix} \lambda \\ \lambda\omega \\ \lambda \omega^2 \\ \vdots \\ \lambda\omega^{n-1} \end{bmatrix}=\lambda \mathbf{v}.$

Since $\mathbf{v}$ is a nonzero vector, it follows that $\lambda$ is an eigenvalue of $A$ and $\mathbf{v}$ is an eigenvector corresponding to $\lambda$.

The above argument holds for any $n$-th root of unity $\omega$.
We take $\omega=\zeta^k$, where $k$ runs from $0$ to $n-1$.
It follows that the vector
$\mathbf{v}_k:=\begin{bmatrix} 1 \\ \zeta^k \\ \zeta^{2k} \\ \vdots \\ \zeta^{k(n-1)} \end{bmatrix}$ is eigenvector corresponding to the eigenvalue
$\lambda_k:=a_0+a_1\zeta^k+a_2\zeta^{2k}+\cdots a_{n-2}\zeta^{k(n-2)}+a_{n-1}\zeta^{k(n-1)}$ for each $k=0,1, \dots, n-1$.

We claim that the vectors $\mathbf{v}_k$ are linearly independent.

To see this, form a matrix whose column vectors are these vectors. That is, we consider
$B=\begin{bmatrix} 1& 1 & 1 & \dots & 1 &1 \\ 1&\zeta & \zeta^{2} & \dots & \zeta^{n-2} & \zeta^{n-1} \\ 1&\zeta^{2} & \zeta^{4} & \dots & \zeta^{2(n-2)} & \zeta^{2(n-1)} \\ 1& \zeta^{3} & \zeta^{6} & \dots & \zeta^{3(n-2)} & \zeta^{3(n-1)} \\ \vdots& \vdots & \vdots & \dots & \vdots & \vdots \\ 1& \zeta^{n-1} & \zeta^{2(n-1)} & \dots & \zeta^{(n-1)(n-2)} &\zeta^{(n-1)(n-1)} \end{bmatrix}.$

This is the Vandermonde matrix and its determinant is
$\det(B)=\prod_{i < j }(\zeta^j-\zeta^i)\neq 0.$ Thus, the matrix $B$ is nonsingular, hence its column vectors are linearly independent. It follows that $\lambda_k$, $k=0, 1, \dots, n$ are all the eigenvalues of $A$. Since the determinant is the product of all the eigenvalues of $A$, we have \begin{align*} \det(A)&=\prod_{k=0}^{n-1}\lambda_k\\ &=\prod_{k=0}^{n-1}(a_0+a_1\zeta^k+a_2 \zeta^{2k}+\cdots+a_{n-1}\zeta^{k(n-1)}), \end{align*} as required. This completes the proof. Add to solve later

### 6 Responses

1. Parthasarathi Mukhopadhyay says:

Kindly clarify why the matrix B is taken as (n-1)X(n-1)? It seems that a column of 1at the beginning and the actual second row is missing.

• Yu says:

Thank you for pointing out this. That was a typo and I modified it. Thank you!!

2. Zubair says:

Respected Doctor,
I don’t think the second entry you computed for “Av” vector is right, because when you take ‘w’ common then the a_n_1 term should be multiplied with just w instead of w^n-1

• Yu says:

Dear Zubair,

Thank you for the comment. I’m not exactly sure what you suggested.

Do you ask why we have $\omega^{n-1]$ in front of $a_{n-1}$? in the following computation?

====================================================

\begin{align*} &a_{n-1}+a_0\omega+\cdots+a_{n-3}\omega^{n-2}+a_{n-2}\omega^{n-1}\\ &=(a_{n-1}\omega^{n-1}+a_0+\cdots+a_{n-3}\omega^{n-3}+a_{n-2}\omega^{n-2})\omega \\ &=(a_0+\cdots+a_{n-3}\omega^{n-3}+a_{n-2}\omega^{n-2}+a_{n-1}\omega^{n-1})\omega \\ &=\lambda \omega \end{align*}

====================================================

This is because we have $\omega^n=1$ since $\omega$ is an $n$-th root of unity.

• Zubair says:

Yes! exactly I was asking this qustion.
Thankyou so much I got your point.

3. Ragib Zaman says:

Let $p(x) = \sum_{i=0}^{n-1} a_i x^i,$ and note that $A=p(X),$ where $X$ is (the transpose of) the companion matrix of $t^n-1.$ The result then follows from the fact that if $\lambda$ is an eigenvalue of $M,$ then $q(\lambda)$ is an eigenvalue of $q(M)$ for polynomial (in fact, analytic) functions $q.$

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