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I reduced a certain determinant to the following which has the form: \begin{vmatrix} 1&1&\cdots & 1 \\ -a_1&a_2&\cdots &a_n\\ -a_1^{2}&a_2^{2}&\cdots&a_n^{2}\\ \vdots&\vdots&\ddots& \vdots\\ -a_1^{n-1}&a_2^{n-1}&\cdots&a_n^{n-1}\\ \end{vmatrix}

To clarify a bit, it is exactly the Vandermonde determinant except the first column is negative, while the $(1,1)$-entry is still $1$.

I think the key is to apply Vandermonde determinant yet I can't proceed. There may be a quick answer to this, however. Any hints?

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    $\begingroup$ Have you tried to developing the determinant using the first column? You can then use the Vandermonde formula for the determinants of the submatrices. $\endgroup$
    – Olivier
    Mar 11, 2017 at 18:35
  • $\begingroup$ Is that you? No that can't be right. Karma works on differential forms in Vegas now. $\endgroup$ Mar 11, 2017 at 18:36
  • $\begingroup$ Wikipedia claims it is $\prod_{0\leq k\leq l\leq n}(a_l- a_k)$ $\endgroup$ Mar 11, 2017 at 18:43
  • $\begingroup$ @mathreadler: I doubt it. As you write it, with $k\leq l$, your expression is zero. If you sharpen it up you've then said it's the Vandermonde determinant, which it is not. $\endgroup$ Mar 11, 2017 at 18:50
  • $\begingroup$ hmm it is not vandermonde determinant? which of them are supposed to be negative? only first column? I don't see any other difference than first column being multiplied by -1. $\endgroup$ Mar 11, 2017 at 18:55

2 Answers 2

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By linearity in or Laplace expansion along the first column, the determinant is equal to $$ \begin{vmatrix} -1&1&\cdots & 1 \\ -a_1&a_2&\cdots &a_n\\ -a_1^{2}&a_2^{2}&\cdots&a_n^{2}\\ \vdots&\vdots&\ddots& \vdots\\ -a_1^{n-1}&a_2^{n-1}&\cdots&a_n^{n-1}\\ \end{vmatrix} + 2\begin{vmatrix} a_2&\cdots &a_n\\ a_2^{2}&\cdots&a_n^{2}\\ \vdots&\ddots& \vdots\\ a_2^{n-1}&\cdots&a_n^{n-1}\\ \end{vmatrix}. $$ Now each of the two summands is a multiple of a Vandermonde determinant.

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  • $\begingroup$ Can you explain a bit? I tried the expansion but how to simplify to yours? $\endgroup$
    – Karma
    Mar 12, 2017 at 4:16
  • $\begingroup$ In a neat way he has replaced your first column by $(-1,-a_1,\dots,-a_1^{n-1})^T+(2,0,\dots,0)^T$ and used linearity. $\endgroup$ Mar 12, 2017 at 7:38
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It is also possible to evaluate a more general determinant, where the clever observation of @user1551 isn't available. Let $c_0,\dots, c_1$ be arbitrary and let $X$ be a polynomial variable. Then we will evaluate $$ \begin{vmatrix} c_0&1&\cdots & 1 \\ c_1 X&a_2&\cdots &a_n\\ c_2 X^2&a_2^{2}&\cdots&a_n^{2}\\ \vdots&\vdots&\ddots& \vdots\\ c_{n-1}X^{n-1}&a_2^{n-1}&\cdots&a_n^{n-1}\\ \end{vmatrix}. $$

By a Lagrange expansion on the first column we get that this is $$ \sum_{0}^{n-1} (-1)^{k} c_k X^k V_k \hspace{10em}\text{(*)} $$ where $V_k$ is the determinant of the corresponding $(n-1)\times(n-1)$ minor in $$ \begin{vmatrix} 1&\cdots & 1 \\ a_2&\cdots &a_n\\ a_2^{2}&\cdots&a_n^{2}\\ \vdots&\ddots& \vdots\\ a_2^{n-1}&\cdots&a_n^{n-1}\\ \end{vmatrix}. $$ But now consider the special case where all $c_k=1$; this is just the standard Vandermonde determinant, so we have $$ \sum_{0}^{n-1} (-1)^{k} X^k V_k = \prod_{r=2}^{n} (X-a_r)\prod_{2\leqslant s < t \leqslant n} (a_t - a_s). $$

We write, in the usual way, $$ \prod_{r=2}^{n} (X-a_r)=\sum_{k=0}^{n-1}(-1)^{k} \sigma_{n-1-k} X^k; $$ these are just the usual symmetric functions of the $a_2,\dots, a_{n}$ taking their sum, their sum two at a time, and so on.

The coefficients $V_k$ then all have a common factor $\prod_{2\leqslant s < t \leqslant n} (a_t - a_s)$, the Vandermonde of $a_2, \dots, a_n$ (which deals with the skew-symmetry in these variables); the other factor in each case is just the symmetric function $\sigma_{n-k}$: $$ V_k= \sigma_{n-k}\prod_{2\leqslant s < t \leqslant n} (a_t - a_s).$$

Taking $c_0=1, c_1=\dots=c_n=1$ and substituting $X=a_1$ in (*) gives the value for the determinant of the original question.

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