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Let $x_i,i\in\{1,\cdots,n\}$ be real numbers, and $s_k=x_1^k+\cdots+x_n^k$, I'm asked to calculate $$ |S|:= \begin{vmatrix} s_0 & s_1 & s_2 & \cdots & s_{n-1}\\ s_1 & s_2 & s_3 & \cdots & s_n\\ s_2 & s_3 & s_4 & \cdots & s_{n+1}\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ s_{n-1} & s_{n} & s_{n+1} & \cdots & s_{2n-2} \end{vmatrix} $$ and to prove that $|S|\ge 0$ for all possible real $x_i$.

I found that $$ |S|=\det[(v_1+\cdots v_n), (x_1v_1+\cdots+x_nv_n),\cdots,(x_1^{n-1}v_1+\cdots+x_n^{n-1}v_n)],\quad\text{where}\, v_j=\begin{bmatrix} 1 \\ x_j \\ \vdots\\ x_j^{n-1} \end{bmatrix} $$ Due to multilinearity of the $\det$ function, I sense it might have something to do with Vandermonde determinant. In fact, it must have the form $$|S|=(\det[v_1,\cdots, v_n])\cdot \text{something}$$ But that "something" involves many cyclic sums and is therefore a horrible mess..

Anyway, is there a neat way to calculate this tricky determinant? Thanks!

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2 Answers 2

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Hint:

$$\begin{bmatrix}1&1&1\\x_1&x_2&x_3\\x_1^2&x_2^2&x_3^2\end{bmatrix}\begin{bmatrix}1&x_1&x_1^2\\1&x_2&x_2^2\\1&x_3&x_3^2\end{bmatrix}=\begin{bmatrix}1+1+1&x_1+x_2+x_3&x_1^2+x_2^2+x_3^2\\x_1+x_2+x_3&x_1^2+x_2^2+x_3^2&x_1^3+x_2^3+x_3^3\\x_1^2+x_2^2+x_3^2&x_1^3+x_2^3+x_3^3&x_1^4+x_2^4+x_3^4\end{bmatrix}$$

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    $\begingroup$ ....I can't believe my eyes but yeah, it's darn good! $\endgroup$
    – Vim
    Sep 4, 2015 at 7:29
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    $\begingroup$ Where does the inspiration come from? I worked out $|S|$ for $n=1$ and $n=2$, and then also for $n=3$ but had a computer factor the result. Seeing the square of the Vandermonde determinant in the $n=3$ case was a strong suggestion to multiply two Vandermonde matrices together. $\endgroup$
    – 2'5 9'2
    Sep 4, 2015 at 7:30
  • $\begingroup$ The proof as written in index notation (with doubled indices summed) is similarly cute. Defining $A_{jk}=x_j^k$, we have $$s_{j+l}=\sum_k x_k^{j+l} = x_k^j x_k^l =A_{jk}A_{lk}=(A A^T)_{jl}$$ $\endgroup$ Dec 22, 2015 at 0:51
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Another way to the formula: I think that you have almost finished the job.

I consider your formula with the $v_k$. Replace them by $w_1,..,w_n$ any vectors in $E=\mathbb{R}^n$. Your formula give then an application $f(w_1,..w_n)$ from $E^n$ to $\mathbb{R}$, and you have see that this is $a{\rm det}(w_1,...,w_n)$ for a constant $a$, depending on the $x_k$, but not on the $w_k$. Now to find $a$, you can take for the $w_k$, the vectors of the canonical basis of $E$, and it is easy to finish.

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