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Prove that $$\begin{vmatrix} 1 & 1 & \cdots & 1 \\ a_0 & a_1 & \cdots & a_n \\ a_0^2 & a_1^2 & \cdots & a_n^2 \\ \vdots & \vdots & \ddots & \vdots \\ a_0^n & a_1^n & \cdots & a_n^n \\\end{vmatrix}=\prod_{j<i}(a_i-a_j)$$. I have trouble with this, I thought it'd be doable with Laplace's theory but I fail to understand it.

jimjim
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Raphael
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2 Answers2

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It's the determinant of a Vandermonde matrix. The standard argument to compute the determinant is to write $$f(x) = \begin{vmatrix} 1 & 1 & \cdots & 1 \\ a_0 & a_1 & \cdots & x \\ a_0^2 & a_1^2 & \cdots & x^2 \\ \vdots & \vdots & \ddots & \vdots \\ a_0^n & a_1^n & \cdots & x^n \\\end{vmatrix}.$$ This is a polynomial in $x$ of degree $n$, and by the properties of determinants (the determinant of a matrix with two columns equal is zero), you have that $a_0, a_1, \dots, a_{n-1}$ are roots. Therefore $$ f(x) = A \cdot (x-a_0) \cdot (x-a_1) \cdot\ldots \cdot (x- a_{n-1}). $$ Now looking at the matrix, you see that the coefficient $A$ of $x^{n}$ is a similar determinant of size one less. Use induction to conclude.

Andreas Caranti
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  • Perhaps I'm missing something basic here, but I think the "by the properties of determinants you have...are the roots" part is almost, or without the almost, as messy as evaluating directly the determinant by minors and induction (say, substracting the first row from all the others and etc.)... – DonAntonio Mar 03 '13 at 13:25
  • @DonAntonio, I was thinking of the fact that a matrix with two columns equal has determinant zero. – Andreas Caranti Mar 03 '13 at 13:26
  • I still can't see it: what two columns? Do you mean substracting some scalar multiple of one of them to another one? – DonAntonio Mar 03 '13 at 13:28
  • @DonAntonio, I mean, if one substitutes $x = a_0$, say, in the definition of $f(x)$ I gave, we are talking of the determinant of a matrix in which the first and the last column are the same. – Andreas Caranti Mar 03 '13 at 13:33
  • Rats, that's easy! Thanks @Andreas and, of course, +1 – DonAntonio Mar 03 '13 at 13:35
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    Thanks @DonAntonio ;-) I will add a few lines to the argument to make it more explicit. – Andreas Caranti Mar 03 '13 at 13:36
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Let $$V(a_0,\ldots,a_n)=\begin{vmatrix} 1 & 1 & \cdots & 1 \\ a_0 & a_1 & \cdots & a_n \\ a_0^2 & a_1^2 & \cdots & a_n^2 \\ \vdots & \vdots & \ddots & \vdots \\ a_0^n & a_1^n & \cdots & a_n^n \\\end{vmatrix}.$$

We have $V(a_0,a_1)=a_1-a_0.$ Suppose $n\geq3$, and we denote $l_{0},l_1,\ldots,l_n$the lines of $V$. We write $$V=\det((l_{0},l_1-a_0l_{0},l_2-a_0l_1,\ldots,l_n-a_0l_{n-1})^T)$$ we find $$V=\begin{vmatrix} 1 & 1 & \cdots & 1 \\ 0 & a_1-a_0 & \cdots & a_n-a_0 \\ 0 & a_1(a_1-a_0) & \cdots & a_n(a_n-a_0) \\ \vdots & \vdots & \ddots & \vdots \\ 0 & a_1^{n-1}(a_1-a_0) & \cdots & a_n^{n-1}(a_n-a_0) \\\end{vmatrix}.$$ So, $$V(a_0,\ldots,a_n)=V(a_1,\ldots,a_n)\prod_{k=1}^n(a_k-a_0).$$ Now, immediate induction provides: $$V(a_0,\ldots,a_n)=\prod_{0\leq i<j\leq n}(x_j-x_i).$$