1
$\begingroup$

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 $N$?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

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.

Improve this answer
$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.