I have an $n\times n$ matrix of the form $A=\begin{bmatrix} 1 & 1 & \dots & 1 \\ x_{1} & x_{2} & \dots & x_{n} \\ x_{1}^2 & x_{2}^2 & \dots & x_{n}^2 \\ \vdots & \vdots & & \vdots \\ x_{1}^{n-1} & x_{2}^{n-1} & \dots & x_{n}^{n-1} \end{bmatrix}$ where I know that every pair of $x_i$ is distinct. I would like to show that $A$ is invertible by showing that it has full rank but I can't find a way to prove that. I don't really need a proof rather than a confirmation whether A is invertible or not and if my idea (about the rank) will work. Thank you!
invertibility of specific matrix through calculation of its rank
0
$\begingroup$
linear-algebra
matrices
-
2Estimating the rank of this will be hard to do. On the other hand, this is a Vandermonde matrix (or a Vandermonde matrix's transpose, see [here](https://en.wikipedia.org/wiki/Vandermonde_matrix)), so the condition on the $x_{i}'s$ is sufficient for invertibility. – 2017-02-24
-
0Got it, thanks a lot! :) – 2017-02-24