Consider a complex matrix $A\in\mathbb{C}^{n+1\times m}$ such that $$ A=\begin{bmatrix} 1 & 1 & \dots & 1\\ Bc_1 & Bc_2 & \dots & Bc_{m} \end{bmatrix}, $$ where $c_1,\dots ,c_m\in\mathbb{C}^p$ are fixed and $B\in\mathbb{C}^{p\times n}$ is such that $$ (Bc_k)_j = \sum_{\ell=1}^{p}(c_k)_\ell e^{i\lambda_{k,\ell} j}, $$ with $\lambda_{k,\ell}\in\mathbb{C}$ for all $k\in \{1,\dots, m\}$ and $\ell\in\{1,\dots ,p\}$. (In the latter expression $j$ is an index and $i$ is the imaginary unit).

I am looking for a closed form of the kernel of the matrix $A$. I think that working on the determinant might be useful.

As a first step, I am working on the special case where $p=1$. In this case, ommiting the $\ell$ subscript from $\lambda$, we have $$ \begin{bmatrix} 1 & 1 & \dots & 1\\ c_1e^{i\lambda_1} & c_2e^{i\lambda_2} & \dots & c_me^{i\lambda_m}\\ c_1e^{i2\lambda_1} & c_2e^{i2\lambda_2} & \dots & c_me^{i2\lambda_m}\\ \vdots & \vdots & &\vdots\\ c_1e^{in\lambda_1} & c_2e^{in\lambda_2} & \dots & c_me^{in\lambda_m}\\ \end{bmatrix}, $$ which is very similar to a Vandermonde matrix. Note that in this case we can, for each column $k\in\{1,\dots ,m\}$, multiply by one by applying a column-multiplication elementary matrix which multiplies by $z_k:=c_ke^{i\lambda_k}$ and an other which multiplies by $z_k^{-1}$. Therefore, by the Laplace expansion over the first row and the Vandermonde structure of the related submatrices one obtains $$ det(A)=\left[\prod_{k=1}^mc_ke^{i\lambda_k}\right]\cdot\left[\sum_{k=1}^m\frac{(-1)^{k+1}}{c_ke^{i\lambda_k}}\prod_{(s_1,s_2)\in S_k}(e^{i\lambda_{s_1}}- e^{i\lambda_{s_2}})\right], $$ where $S_k=\{(s_1,s_2)\in [m]^2:s_1<s_2, s_1\neq k, s_2\neq k\}$. Even though this is a closed form expression, it is not precisely informative for the kernel of $A$.

Any comments or suggestions will be greatly appreciated.

New contributor
PIII is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
  • $\begingroup$ A similar problem is discussed in mathoverflow.net/questions/236323, see also arxiv.org/abs/2103.10776 (J. Phys. A: Math. Theor. 54, 375201 (2021)). $\endgroup$
    – Fred Hucht
    2 days ago
  • $\begingroup$ Thank you, Fred, for your comment. I have read your post and the Appendix B of your paper, it is very insightful. I see that you're able to reduce the problem of computing the determinant of the Vandermonde type matrix into the problem of computing the determinant of a Hankel matrix. However, unfortunately I don't see how to obtain a benefit by doing something similar in my case. How would you use that to have an explicit form for the kernel of the matrix A? $\endgroup$
    – PIII


Your Answer

PIII is a new contributor. Be nice, and check out our Code of Conduct.

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.