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I'm trying to reduce the following matrix given distinct complex numbers $\beta_1, \beta_2,...,\beta_N$ with the knowledge that $N \geq M+1$.

From my testing I can see that this reduces to an identity matrix with $N=M+1$, and if $N > M +1$ any rows after row $M+1$ become all zeroes.

But I'm unsure of how to fully show this in a convincing argument.

$\begin{bmatrix} (\beta_1)^M& (\beta_1)^{M-1}& ... & \beta_1 & 1\\ (\beta_2)^M & (\beta_2)^{M-1} & ... & \beta_2 & 1 \\ \vdots &\vdots & &\vdots & 1 \\ (\beta_N)^M & (\beta_N)^{M-1} & ... & \beta_N & 1 \ \end{bmatrix}$

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Consider first the case where $N = M + 1$, and let $A$ be the given matrix. Further, let $B_N$ be the matrix with $1$s in the off-diagonal, and $0$s elsewhere, e.g. $$B_3 = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}.$$ Then $$AB_N = \begin{bmatrix} 1 & \beta_1 & \cdots & \beta_1^M \\ 1 & \beta_2 & \cdots & \beta_2^M \\ \vdots & \vdots & \ddots & \vdots \\ 1 & \beta_N & \cdots & \beta_N^M \end{bmatrix}$$ is a Vandermonde Matrix, and is invertible if and only if $\beta_1, \beta_2, \ldots, \beta_N$ are all distinct. Since $B_N$ is its own inverse, it follows that $A$ is also invertible. Hence, row reduction will yield the identity matrix, as conjectured.

Now, if $N \ge M + 1$, then we can simply ignore rows $M+2, M+3, \ldots, N$, and row reduce the first $M + 1$ rows. By the above argument, they will also reduce down to the identity matrix. Now that a pivot exists in each column, one can easily reduce the rows below to $0$ rows.

So, in conclusion, all your conjectures are true.

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