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Consider a block Vandermonde square matrix

$$X = \begin{bmatrix} \mathbf{I} & \mathbf{\Omega}(i_0) & \mathbf{\Omega}(2i_0) & \ldots & \mathbf{\Omega}((2p-1)i_0) \\ \mathbf{I} & \mathbf{\Omega}(-i_0) & \mathbf{\Omega}(-2i_0) & \ldots & \mathbf{\Omega}(-(2p-1)i_0) \\ \mathbf{I} & \mathbf{\Omega}(i_1) & \mathbf{\Omega}(2i_1) & \ldots & \mathbf{\Omega}((2p-1)i_1) \\ \mathbf{I} & \mathbf{\Omega}(-i_1) & \mathbf{\Omega}(-2i_1) & \ldots & \mathbf{\Omega}(-(2p-1)i_1) \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ \mathbf{I} & \mathbf{\Omega}(i_{p-1}) & \mathbf{\Omega}(2i_{p-1}) & \ldots & \mathbf{\Omega}((2p-1)i_{p-1}) \\ \mathbf{I} & \mathbf{\Omega}(-i_{p-1}) & \mathbf{\Omega}(-2i_{p-1}) & \ldots & \mathbf{\Omega}(-(2p-1)i_{p-1}) \end{bmatrix} \in \mathbb{R}^{4p \times 4p}$$ where $$\mathbf{\Omega}(i) = \begin{bmatrix} \cos(2\pi i /M) &-\sin(2\pi i/M) \\ \sin(2\pi i /M) &\cos(2\pi i /M) \end{bmatrix}$$ which is a rotation matrix, $0<i_0 < i_1 < \ldots < i_{p-1} <M$

Note that all the entry is $2 \times 2$.

Question

How to show $X$ is invertible?

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  • $\begingroup$ I suspect there are some row/column operations that turn $X$ into a complex Vandermonde matrix (as opposed to a real block-Vandermonde matrix) which is then easily shown to be invertible. $\endgroup$ – JimmyK4542 Jan 3 at 1:04

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