Operator norm of some type of discrete Fourier matrix

Question

Let $N$ be a natural number and let $w$ be a complex number.

We define the $N\times N$ matrix $C_w=(a_{k,l})_{k,l=1}^N$ as follows,
$$ a_{k,l}=\begin{cases}1 & l=k+1\\ w & k=N, l=1\\ 0 & \text{o.w} \end{cases} $$ The characteristic polynomial of $C_w$ is just $p(\lambda) = \lambda^N - w $. Thus, the eigenvalues of $ C_w $ are just the $N$-th roots of $w$ which are as follows: \begin{equation} \lambda_k = |w|^{1/N} e^{2\pi i k / N} \quad \text{for} \quad k = 0, 1, \ldots, N-1 \end{equation} The $k$-th column of the following matrix is the eigenvector of $C_w$ corresponding to $\lambda_{k-1}$. \begin{equation} F_w=\begin{bmatrix} 1 & 1 & \cdots & 1 \\ \lambda_0 & \lambda_1 & \cdots & \lambda_{N-1} \\ \vdots & \vdots & \cdots & \vdots \\ \lambda_0^{N-1} & \lambda_1^{N-1} & \cdots & \lambda_{N-1}^{N-1} \end{bmatrix} \end{equation}

If $w=1$, then $F_w$ is the discrete Fourier matrix.

Q. Is it possible to formulate the operator norm of the matrix $F_w$ in terms of $w$ and $N$?

Answer 1

You can use the $F_wF_w^*$ argument to calculate the operator norm of the matrix. In fact we have that the elements of $ F_w $ are of the form $f_{kn} = \lambda^k_n $. So the entries of $F_wF_w^*$ are $$ \sum_{r=0}^{N-1} f_{kr}\overline{f_{nr}} = \sum_{r=0}^{N-1} \lambda^k_r\overline{\lambda^n_r} =|w|^{\frac{k+n}{N}} \sum_{r=0}^{N-1} \exp(\frac{2\pi i r (k-n) }{N}) = N|w|^\frac{k+n}{N} \delta_{kn}. $$

This is a diagonal matrix so its norm is the largest modulus on the diagonal. Hence $$ \Vert F_w F_w^* \Vert = \begin{cases} N, \,\, \text{when} \,\, |w| \leq 1 \\ N|w|^\frac{N-2}{N}, \,\, \text{when} \,\, |w| \geq 1 \end{cases} = \Vert F_w \Vert^2. $$ The last equality be the $C^*$ identity.