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 $j^2=-1$.


  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.


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



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