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Currently, most engineering texts (and webpages including Wikipedia) define forward discrete Fourier transform with a negative sign on the exponential. This is a convention and the inverse discrete Fourier transform has a positive sign. It is not uncommon to see the opposite convention in older literature. However, I am interested in the diagonalization of a circulant matrix and the labeling of its eigenvalues is little bit confusing even in modern texts. I am searching Google Books. I thought the negative sign convention on the exponent in DFT is currently practiced but many latest books appear to be inconsistent in labeling eigenvalues of a circulant. What is the accepted practice among mathematicians (I am coming from chemistry background)?

If we diagonalize a circulant matrix, some books say, the eigenvalues of a circulant correspond to the inverse Fourier transform of the row of the circulant matrix. Whereas, many other lecture notes on the circulant from top universties and other books say that the eigenvalues of a circulant correspond to the discrete Fourier transform.

One example, one book "An Introduction to Computational Stochastic PDEs" says,

Lemma 6.44 Let $W$ be the $N \times N$ Fourier matrix (see Definition 1.95) with entries $w_{i j}=\frac{1}{\sqrt{N}} \omega^{(i-1)(j-1)}$ for $\omega:=\mathrm{e}^{-2 \pi \mathrm{i} / N}$. For $\boldsymbol{c} > \in \mathbb{R}^N, W \boldsymbol{c}$ and $W^* \boldsymbol{c}$ are Hermitian. For a Hermitian vector $\boldsymbol{c} \in \mathbb{C}^N, W > \boldsymbol{c}$ and $W^* \boldsymbol{c}$ are real.

The eigenvalues of a circulant matrix are given by $N$ times the inverse DFT of the first column and the eigenvectors are the columns of the Fourier matrix $W$. This leads to the following factorisation of a circulant matrix.

Another one states

3.2 Matrix Operations on Circulant Matrices The following theorem summarizes the properties derived in the previous section regarding eigenvalues and eigenvectors of circulant matrices and provides some easy implications.

Theorem 3.1. Every circulant matrix $C$ has eigenvectors $y^{(m)}=$ $\frac{1}{\sqrt{n}}\left(1, e^{-2 \pi i m / n}, \cdots, e^{-2 \pi > i(n-1) / n}\right)^{\prime}, m=0,1, \ldots, n-1$, and corresponding eigenvalues $$ \psi_m=\sum_{k=0}^{n-1} c_k e^{-2 \pi i m k / n} $$ and can be expressed in the form $C=U \Psi U^*$, where $U$ has the eigenvectors as columns in order and $\Psi$ is $\operatorname{diag}\left(\psi_k\right)$. In particular all circulant matrices share the same eigenvectors, the same matrix $U$ works for all circulant matrices, and any matrix of the form $C=U \Psi U^*$ is circulant.

This book implies that eigenvalues of a circulant are discrete Fourier transform not the inverse DFT.

Another one: This one clearly says so.

Example B.4 The real matrices $\boldsymbol{A}$ and $\boldsymbol{B}$ below are Toeplitz and circulant, respectively. $$ \boldsymbol{A}=\left[\begin{array}{lll} 2 & 5 & 1 \\ 4 & 2 & 5 \\ 3 & 4 & 2 \end{array}\right] \quad \boldsymbol{B}=\left[\begin{array}{ccc} 2 & 5 & 1 \\ 1 & 2 & 5 \\ 5 & 1 & 2 \end{array}\right] $$ If $\boldsymbol{A}$ is an $N \times N$ circulant matrix, then the following holds:

  • The eigenvectors of $\boldsymbol{A}$ equal the columns of the Fourier matrix $\boldsymbol{U}$ in (B.7).
  • The eigenvalues of $\boldsymbol{A}$ equal the entries of $\boldsymbol{U}^* \boldsymbol{a}$ where $\boldsymbol{a}$ is any column of $\boldsymbol{A}$. Hence, the eigenvalues of a circulant matrix are directly the DFT of any of the columns (or rows) of that matrix [139].

Reference 139 is a a review paper on Toeplitz and Circulants.

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Q: Are the eigenvalues of a circulant matrix given by the DFT or Inverse DFT of a column of matrix elements?

A: It makes no difference, the list $f$ of Fourier coefficients and the list $f'$ of inverse Fourier coefficients of a list of $N$ complex numbers are identical up to a permutation. (This is true simply because $e^{2\pi i kn/N}=e^{-2\pi i k'n/N}$ with $k'=N-k$.)

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  • $\begingroup$ Thanks, could you clarify the term "identical up to a permutation"? $\endgroup$
    – ACR
    Commented Jan 11, 2023 at 14:35
  • $\begingroup$ Basically, if we have a sequence of real numbers (=signal), whether we take a discrete Fourier transform of that sequence with a plus sign or a minus sign on the exponential, the output will be complex conjugate of each other. $\endgroup$
    – ACR
    Commented Jan 11, 2023 at 16:11
  • $\begingroup$ the statement that $f$ and $f'$ are identical up to a permutation (meaning, a reordering of the elements in the list) also holds if the circulant has complex matrix elements $\endgroup$ Commented Jan 11, 2023 at 16:41

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