# Condition number on the DFT-like complex vandermonde matrix

Given $$M \in \mathbb{N}$$ and $$0 < L \le M$$, $$L \in \mathbb{N}$$ consider a set of $$L-1$$ integers, such that $$0 \le i_1 < i_2 \ldots < i_{L-1} \le M$$

Note that this index set has symmetry inside because they are generated from FFT on a real signal using $$M$$ equ-spaced sampling. For example, so let's say if $$i_1= 1$$, then $$i_{L-1}$$ must be $$M-1$$.

Then we have the following complex Vandermonde matrix based on the above $$\mathbf{A} \triangleq V(e^{2\pi j i_1/M},\ldots,e^{2\pi j i_{L-1}/M}) = \begin{bmatrix} 1 & e^{2\pi j i_1/M} & e^{4\pi j i_1/M} & \ldots e^{2(L-1)\pi j i_1/M}\\ 1 & e^{2\pi j i_2/M} & e^{4\pi j i_2/M} & \ldots e^{2(L-1)\pi j i_2/M}\\ 1 & \ldots & \ldots & \ldots \\ 1 & e^{2\pi j i_{L-1}/M} & e^{4\pi j i_{L-1}/M} & \ldots e^{2(L-1)\pi j i_{L-1}/M} \end{bmatrix} \in \mathbb{C}^{L\times L}$$ where $$j^2=-1$$.

## Observation:

1. If L = M + 1, then $$\mathbf{A}$$ is the classical DFT matrix, which has perfect conditioning up to scaling a constant. So the problem here can be thought as DFT-like vandermonde matrix.

2. Fixing $$i_1,\ldots, i_{L-1},$$I found the $$cond(\mathbf{A})$$ grows with increasing M, which makes sense because each row would approaching $$\begin{bmatrix} 1 & 1 & 1& \ldots & 1 \end{bmatrix}$$ as $$M$$ increasing.

## Question

I want to know the any useful bound on the condition number of $$\mathbf{A}$$.