We are given the following information:- Let $z,w \in \mathbb{C^n}$ be signals, $z_{\delta} = \Omega_nz$ and $w_{\delta} = \Omega_nw$ be the discrete Fourier transformation of $z$ and $w$. We define $\Omega_n$ as the following:
$$\Omega_n = \frac{1}{\sqrt n} \begin{bmatrix}1&1&1&1&...&1\\1&\omega&\omega^2&\omega^3&...&\omega^{n-1}\\1&\omega^2&\omega^4&\omega^6&...&\omega^{2(n-1)}\\1&\omega^3&\omega^6&\omega^9&...&\omega^{3(n-1)}\\...&...&...&...&...&...\\1&\omega^{n-1}&\omega^{2(n-1)}&\omega^{3(n-1)}&...&\omega^{(n-1)(n-1)}\end{bmatrix}$$
Given the fact that the columns of the DFT matrix are normalized, which of the following statements are true?
- We can say that $z_{{\delta}_{n-1}}$ describes the share of the highest frequencies of of the signal $z$.
- In the rows of the DFT matrix $\Omega_n$, you will find the eigenvectors in a circulant matrix.
- The convolution of $z$ and $w$ correspond to the elemental multiplication of the transformed signals: $z \cdot w= z_{\delta}⋅w_{\delta}$
For $2.$ I'm pretty sure that it's false because when we take the case of a four point clockwise DFT matrix, we can see that the rows don't contain the eigenvectors in a circulant matrix. For $3.$ I have a feeling that it's true but I'm not sure if I can explain it.