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Given postive integer $M \in \mathbb{N}$, find the condition on the the combination of $distinct$ $p$ integers $$i_{0}, i_{1}, \ldots, i_{p-1} \in \{0,1,\ldots,M-1 \}, \omega = e^{-j\frac{2\pi}{M}}$$ such that $$ \begin{bmatrix} \omega^{i_0} \\ \omega^{i_1} \\ \omega^{i_2} \\ \vdots \\ \omega^{i_{p-1}} \end{bmatrix} = \begin{bmatrix} 1 & \omega^{-i_0} & \ldots & \omega^{-(p-1)i_0}\\ 1 & \omega^{-i_1} & \ldots & \omega^{-(p-1)i_1}\\ 1 & \omega^{-i_2} & \ldots & \omega^{-(p-1)i_2}\\ \vdots & \vdots & \vdots & \vdots \\ 1 & \omega^{-i_{p-1}} & \ldots & \omega^{-(p-1)i_{p-1}} \end{bmatrix} K $$ has real solution $K \in \mathbb{R}^{p}$.

What I have found

  1. when $ p = M$, there is always a real solution and the real solution is $\begin{bmatrix} 0& 0 \ldots & 1\end{bmatrix}$
  2. when $p < M$, I can just naively use the rank test in real matrix form, but, is there any better result we can obtain?

Update:

For $p<M$, and for particularly searching for zero-end solution, like $\begin{bmatrix} 0& 0 \ldots & 1\end{bmatrix}$, I have found a systematic way to do so.

But in practice, there are no-zero end solution which makes the problem harder.

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