Fourier transform of a Kronecker product

Question

Let $v=v(\theta)$ and $w=w(\theta)$ vectors with $d$ components defined as \begin{align} v&=(e^{i\theta}\quad e^{i2\theta}\quad \ldots\quad e^{id\theta})\\ w&=(e^{i\theta(\rho_1-1)}\quad e^{i\theta(\rho_2-2)}\quad \ldots\quad e^{i\theta(\rho_d-d)}), \end{align} For $\rho_i$ non-negative integers. Now, write the Fourier transform of the Kronecker product $v^T\otimes w$ $$\int (v^T(\theta)\otimes w(\theta)) e^{-i\theta r} d\theta.$$ There is a way to factorize the previous integral in terms of a Kronecker product of the Fourier transforms of the vectors?

Answer 1

Let's try it out. Using the short-hand notation ${\mathcal F}\{\cdot\} = \int (\cdot) {\rm e}^{-\jmath \theta r}{\rm d}\theta$, for one component of your vectors we have $${\mathcal F}\{{\rm e}^{\jmath \theta k}\} = \int {\rm e}^{\jmath \theta k} {\rm e}^{-\jmath \theta r} {\rm d}\theta = \delta (k-r)$$ (skipping the integration constant here for clarity). Therefore $$\begin{align} {\mathcal F}\{v(\theta)\} & = [\delta(1-r), \ldots, \delta(d-r)] \\ {\mathcal F}\{w(\theta)\} & = [\delta(\rho_1-1-r), \ldots, \delta(\rho_d-r)] \\ {\mathcal F}\{v^T(\theta) \otimes w(\theta)\} & = \begin{bmatrix} \delta(n\cdot (\rho_m-m) - r) \end{bmatrix}_{n=1,2,\ldots, d, m=1, 2, \ldots, d} \\ {\mathcal F}\{v^T(\theta)\} \otimes {\mathcal F}\{w(\theta)\} & = \begin{bmatrix} \delta(n - r)\cdot\delta( (\rho_m-m) - r) \end{bmatrix}_{n=1,2,\ldots, d, m=1, 2, \ldots, d} \end{align}$$ using $A = [a_{mn}]_{m=1,2,...M,n=1,2,...,N}$ to define the elements of an $M \times N$ matrix.

Quite clearly, the two are not the same... so my answer would be no.