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Let $\{\alpha_{i}\}_{i=1}^{n}$ be distinct numbers. What is the determinant of the $n$ by $n$ matrix \begin{gather} \begin{pmatrix} \alpha_{1}^{n-1} & \alpha_{1}^{n-2} & \cdots & 1 \\ \alpha_{2}^{n-1} & \alpha_{2}^{n-2} & \cdots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ \alpha_{n}^{n-1} & \alpha_{n}^{n-2} & \cdots & 1 \end{pmatrix} \end{gather} ?

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marked as duplicate by user1551 linear-algebra Aug 25 '15 at 11:12

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    $\begingroup$ Do you mean Vandermonde matrix? $\endgroup$ – GAVD Aug 25 '15 at 10:49
  • $\begingroup$ Oh, i didn't know :) thanks @GAVD $\endgroup$ – BasicUser Aug 25 '15 at 10:50
  • $\begingroup$ However, the columns aren't in the same order, so i guess there might i some cases be a minus sing? $\endgroup$ – BasicUser Aug 25 '15 at 10:52
  • $\begingroup$ What is the easiest, or rather most elemntary way of proving the the product formula? $\endgroup$ – BasicUser Aug 25 '15 at 10:58
  • $\begingroup$ @BasicUser if $P$ is a permutation matrix that switch two columns, then $\det(P)=-1$. Now if you have $n$ columns, you need $\nu=\lfloor n/2\rfloor$ switch $P_1,\ldots,P_\nu$ to transform your matrix in a Vandermonde matrix. The you can use $\det(V)=\det(AP_1\ldots P_{\nu})=\det(A)(-1)^{\nu}$. $\endgroup$ – Surb Aug 25 '15 at 10:59