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.
No comments:
Post a Comment