그냥 흥미로운 사실 하나

수업을 듣다가, 꽤나 흥미로운 사실인 것 같아서 그냥 정리해둡니다.

 

Proposition. Define elementary symmetric polynomials of $n$ variables $\Lambda = \{ \lambda_{1}, \cdots, \lambda_{n} \}$ by
\begin{align*} c_{0} = 1, \quad c_{k} = \sum_{\substack{X \subset \Lambda \\ |X| = k}} \prod_{\lambda \in X} \lambda \quad (1 \leq k \leq n) \end{align*}
and similarly we define
\begin{align*} s_{k} = \sum_{j=1}^{n} \lambda_{j}^{k} \quad (k \geq 0). \end{align*}
Then the set of values $\{c_{0}, c_{1}, \cdots, c_{n} \}$ completely determines $\{ s_{0}, s_{1}, \cdots, s_{n} \}$ and vice versa.

 

Proof. For simplicity, let $c_{k} = 0$ for $k > n$. It is easy to confirm that \begin{align*} (1 - t\lambda_{1}) \cdots (1 - t\lambda_{n}) &= \sum_{k=0}^{\infty} (-1)^{k} c_{k} t^{k}. \end{align*} Similarly, we have \begin{align*} \frac{1}{1 - t\lambda_{1}} + \cdots + \frac{1}{1 - t\lambda_{n}} &= \sum_{k=0}^{\infty} s_{k} t^{k}. \end{align*} Now let \begin{align*} f(x) = (x - t\lambda_{1}) \cdots (x - t\lambda_{n}) = \sum_{k=0}^{\infty} (-1)^{k} c_{k} t^{k} x^{n-k}. \end{align*} On the one hand, considering the log-derivative of $f(x)$, it is easy to confirm that \begin{align*} f'(1) &= (1 - t\lambda_{1}) \cdots (1 - t\lambda_{n}) \left( \frac{1}{1 - t\lambda_{1}} + \cdots + \frac{1}{1 - t\lambda_{n}} \right) \\ &= \sum_{k=0}^{\infty} \left( \sum_{j=0}^{k} (-1)^{j} c_{j} s_{k-j} \right) t^{k}. \end{align*} On the other hand, \begin{align*} f'(1) = \sum_{k=0}^{\infty} (-1)^{k} (n-k) c_{k} t^{k}. \end{align*} Thus it follows that for any $k$ we have \begin{align*} \sum_{j=0}^{k} (-1)^{j} c_{j} s_{k-j} = (-1)^{k} (n-k) c_{k}. \end{align*} Thus if $(c_{k})$ are given, this forms a linear system of $n$ equations for $n$ variables $(s_{k})$ whose corresponding matrix is triangular with non-zero scalar diagonals. Thus we can solve this in terms of $(s_{k})$ and vice versa. This completes the proof.