Is this a circulant matrix?

Question

It's symmetric, but I'm not sure whether it is circulant. In a question that I had asked on MSE a couple of weeks ago, several commenters had said that this is a circulant matrix, and to study the explicit formulas to find its eigenvalues and eigenvectors.

However, I was reading up on circulant matrices again last night and noticed that the "constant diagonals" are actually from left to right, at least according to Wikipedia -- although the page does mention that circulant matrices can be defined in other ways, with different shifts in direction.

This matrix has constant diagonal, but from right to left. Is this still a circulant matrix, and hence the formulas to compute its eigenvalues and eigenvectors are still the same?

Thanks,

$$ A= \begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \\ \end{bmatrix} $$

Answer 1

This is called an anticirculant matrix, which is a special case of Hankel matrix. The eigenvalue/eigenvector formula for circulant matrix does not apply.

Answer 2

Let $\mathbf A = \left\{a_{ij}\right\}_{i,j=0}^n \in \mathbb C^{(n+1) \times (n+1)}$ be defined as $$ a_{ij} = c_{i+j \operatorname{mod} (n+1)}, $$ so $$ \mathbf A = \begin{pmatrix} c_0 & c_1 & \dots & c_n\\ c_1 & c_2 & \dots & c_0\\ \vdots & \vdots & \ddots & \vdots\\ c_n & c_0 & \dots & c_{n-1} \end{pmatrix}. $$ Let's assume every index is taken modulo $n+1$. Then for $\mathbf y = \mathbf A \mathbf x$ $$ y_m = \sum_{j=0}^n a_{mj} x_j = \sum_{j = 0}^n c_{m+j} x_j $$

Let's use the following discrete Fourier transform $$ \mathcal{F}[\mathbf y]_k = \sum_{m=0}^n y_m \exp\left\{-2\pi i\frac{k m}{n+1}\right\} $$ Then $$ \mathcal{F}[\mathbf y]_k = \sum_{m=0}^n \sum_{j=0}^n c_{m+j}x_j \exp\left\{2\pi i\frac{k j}{n+1}\right\}\exp\left\{-2\pi i\frac{k (m+j)}{n+1}\right\} = \\ = \sum_{j=0}^n x_j \exp\left\{2\pi i\frac{k j}{n+1}\right\} \sum_{m=0}^n c_{m+j} \exp\left\{-2\pi i\frac{k (m+j)}{n+1}\right\} = \\ =\sum_{j=0}^n x_j \exp\left\{2\pi i\frac{k j}{n+1}\right\} \sum_{m=-j}^{n-j} c_{m+j} \exp\left\{-2\pi i\frac{k (m+j)}{n+1}\right\} =\\ =\sum_{j=0}^n x_j \exp\left\{2\pi i\frac{k j}{n+1}\right\} \sum_{m'=0}^{n} c_{m'} \exp\left\{2\pi i\frac{k m')}{n+1}\right\} = \\ =\sum_{j=0}^n x_j \exp\left\{2\pi i\frac{k j}{n+1}\right\}\mathcal{F}[\mathbf c]_{k} = \mathcal{F}^{*}[\mathbf x^*]_k \mathcal{F}[\mathbf c]_{k} $$ Thus for $\mathbf A\mathbf x = \lambda \mathbf x$ $$ \mathcal{F}[\mathbf c]_{k} \mathcal{F}^{*}[\mathbf x^*]_k = \lambda\mathcal{F}[\mathbf x]_k $$ Let for simplicity $C_k = \mathcal{F}[\mathbf c]_{k}$. $$ C_{k} \mathcal{F}^{*}[\mathbf x^*]_k = \lambda\mathcal{F}[\mathbf x]_k $$ Recall that discrete Fourier transform can be written in form $$ \mathcal{F}[\mathbf x]_k = (\mathbf F\mathbf x)_k $$ with $\mathbf F^{-1} = \frac{1}{n+1}\mathbf F^H$. Using $\mathbf C = \operatorname{diag} C_k$ one can write the eigenproblem as $$ \mathbf C \mathbf F^* \mathbf x = \lambda \mathbf F \mathbf x $$

The eigenvalues satisfy $$ \operatorname{det}(\mathbf C \mathbf F^* - \lambda \mathbf F) = 0\\ \operatorname{det}(\mathbf C - \lambda\frac{1}{n+1} \mathbf F\mathbf F^\top) = 0 $$ The $\frac{1}{n+1} \mathbf F\mathbf F^\top$ has the following form $$ \frac{1}{n+1} \mathbf F\mathbf F^\top = \begin{pmatrix} 1 & 0 & \dots & 0 & 0\\ 0 & 0 & \dots & 0 & 1\\ 0 & 0 & \dots & 1 & 0\\ \vdots & \vdots & \cdot & \vdots & \vdots\\ 0 & 1 & \dots & 0 & 0 \end{pmatrix} $$ and the eigenvalues (except for $\lambda = C_0$) are the solutions to the problem $$ \left| \mathbf C_\text{AC} - \lambda \mathbf J_n \right| = 0\\ \left| \mathbf J_n \mathbf C_\text{AC} - \lambda \mathbf I_n \right| = 0 $$ where $\mathbf J_n$ is an exchange matrix of size $n$. We arrived to a problem of finging the eigenvalues of the antidiagonal matrix $\mathbf J_n \mathbf C_\text{AC}$.

Next, following this post one can note that $$ (\mathbf J_n \mathbf C_\text{AC})^2 = \operatorname{diag} \begin{pmatrix}C_1 C_n & C_2 C_{n-1} & \dots & C_n C_1\end{pmatrix} $$ so eigenvalues of the original problem are $$ \lambda_0 = C_0\\ \lambda_{i, n-i+1} = \pm \sqrt{C_i C_{n-i+1}},\quad i = 1, 2, \dots $$

Answer 3

Circulant Matrix will have a constant(same) diagonal entries.