Let $p(x_1,x_2,\dots, x_n)=\prod \limits_{i<j}(x_j-x_i)$ and $\sigma$ - some permutation of a set $\{1,2,\dots,n\}$. Prove that $$p(x_{\sigma(1)},x_{\sigma(2)},\dots, x_{\sigma(n)})=\text{sgn}(\sigma)p(x_1,x_2,\dots, x_n).$$ Can anyone give a proof to that fact. Unfortunately any ideas how to solve it.
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asked Feb 2, 2016 at 11:06
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$\begingroup$ Try to prove it for a transposition, by checking all the possible cases. $\endgroup$– Arnaud D.Feb 2, 2016 at 12:09
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1 Answer
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Consider the Vandermonde matrix,
$$V=\begin{bmatrix} 1 & x_{1} & x_{1}^2 & \dots & x_{1}^{n-1} \\ 1 & x_{2} & x_{2}^2 & \dots & x_{2}^{n-1} \\ \vdots \\ 1 & x_{n} & x_{n}^2 & \dots & x_{n}^{n-1} \end{bmatrix} $$
The determinant of the Matrix is $$\det{V}=\prod \limits_{i<j}(x_j-x_i)$$
Now if you change the rows according to the permutation $\sigma$ the determinant matrix $V'$ will be,
$$V=\begin{bmatrix} 1 & x_{\sigma (1)} & x_{\sigma (1)}^2 & \dots & x_{\sigma (1)}^{n-1} \\ 1 & x_{\sigma(2)} & x_{\sigma(2)}^2 & \dots & x_{\sigma(2)}^{n-1} \\ \vdots \\ 1 & x_{\sigma(n)} & x_{\sigma(n)}^2 & \dots & x_{\sigma(n)}^{n-1} \end{bmatrix} $$ The determinant of the new matrix will be, $$\det{V'}=\text{sgn}(\sigma)\det{V}$$ This is because if you interchange the rows an odd number of times the determinant becomes negative of it's initial value and if you change it an even number of times the determinant remains the same.
