Fourier transform of time series by diagonalising matrix

Question

Does there exist a case where the Fourier transform of a time series is found via diagonalising a matrix? Ideally, I am looking for cases where the eigenvalues correspond to frequencies.

Any leads are helpful, thank you!

Answer 1

One important context in which eigenvalues correspond to frequencies of a "Fourier transform" of sorts is the case of a circulant matrix $$ C = \begin{bmatrix} c_0 & c_{n-1} & \cdots & c_2 & c_1 \\ c_1 & c_0 & c_{n-1} & & c_2 \\ \vdots & c_1 & c_0 & \ddots & \vdots \\ c_{n-2} & & \ddots & \ddots & c_{n-1} \\ c_{n-1} & c_{n-2} & \cdots & c_1 & c_0 \\ \end{bmatrix}. $$ In this case, the eigenvalues of $C$ are precisely the discrete Fourier transform of the vector $c = (c_0,c_1,\dots,c_{n-1})$. Notably, the operator associated with $C$ is the discrete circular convolution operator $x \mapsto c*_n x$.

If we want our "eigenvalues" to correspond to the full Fourier transform of a function, we need to move into the infinite dimensional setting. For a function $f$ for which $\int_{-\infty}^\infty |f(x)|^2\,dx$ is finite, we can consider the convolution operator defined so that $$ T_f(g)(t) = (f*g)(t) = \int_{-\infty}^\infty f(\tau) g(t-\tau) d\tau. $$ The eigenvalues of this operator consist precisely of the range of the Fourier transform $\hat f$ of $f$. Similarly, if we consider the operator $T_f$ over the space of periodic functions on $[-\pi,\pi]$ given by $$ T_f(g)(t) = (f\hat *g)(t) = \int_{-\pi}^\pi f(\tau) g(t-\tau) d\tau, $$ Then the eigenvalues of $T_f$ consists precisely of the values appearing in the Fourier series of $f$.