I was recently trying to prove that a Vandermonde type of matrix is full rank and a quick google search did not give me the the proof I had in mind. So here it is,
Consider a matrix of the form,
\begin{equation}
\begin{bmatrix}
1& \alpha_1& \alpha_1^2& \ldots& \alpha_1^{r-1}\\
1& \alpha_2& \alpha_2^2& \ldots& \alpha_2^{r-1}\\
\vdots \\
1& \alpha_r& \alpha_r^2& \ldots& \alpha_r^{r-1}\\
\end{bmatrix} = [\mathbf{a_0} \ldots \mathbf{a_{r-1}}]
\end{equation}
If this matrix is not full rank then there exists coefficients $c_0, c_1, c_2, \ldots, c_{r-1}$ such that $\sum_i c_i \mathbf{a_i} = \mathbf{0}$. Therfore there exist $r$ roots $\alpha_1 \ldots \alpha_r$ of the polynomial $\sum c_i x^i$ of degree less than $r$.
Consider a matrix of the form,
\begin{equation}
\begin{bmatrix}
1& \alpha_1& \alpha_1^2& \ldots& \alpha_1^{r-1}\\
1& \alpha_2& \alpha_2^2& \ldots& \alpha_2^{r-1}\\
\vdots \\
1& \alpha_r& \alpha_r^2& \ldots& \alpha_r^{r-1}\\
\end{bmatrix} = [\mathbf{a_0} \ldots \mathbf{a_{r-1}}]
\end{equation}
If this matrix is not full rank then there exists coefficients $c_0, c_1, c_2, \ldots, c_{r-1}$ such that $\sum_i c_i \mathbf{a_i} = \mathbf{0}$. Therfore there exist $r$ roots $\alpha_1 \ldots \alpha_r$ of the polynomial $\sum c_i x^i$ of degree less than $r$.
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