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Let $x_1, x_2, ... , x_n$ be real numbers, and $A=(a_{ij})$ be the n by n matrix whose entries are $a_{ij}=(x_i)^j$. Evaluate the determinant of A.

Rikka
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1 Answers1

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It is something like Vandermonde matrix. It seem to me that the value of it if neither of the entries is 0, should be equal $\prod_{k=1}^{n}x_{k}\prod_{1\leq i<j\leq n}(x_{j}-x_{i})$.

kolobokish
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  • Would you please explain a little bit? – Rikka Nov 10 '16 at 02:13
  • Yes, of course. First of all you can look up Vandermonde matrix, for example here https://en.wikipedia.org/wiki/Vandermonde_matrix . So the Vandermonde matrix is alike yours, but the entries are a kind of $x_{j}^{i-1}$, and its determinant is $\prod_{i \neq j} (x_{j}-x_{i})$. So to obtain your determinant, we should pull out $x_{j}^{1}$, from each row. And of course if we can do so(i.e. $x_{j}\neq 0 $), then the determinant will have the above form (i.e Vandermonde determinant multiplied by first entries). The form of Vandermonde determinant can be obtained by induction. (the only way I know). – kolobokish Nov 10 '16 at 02:19