그냥 흥미로운 사실 하나

수업을 듣다가, 꽤나 흥미로운 사실인 것 같아서 그냥 정리해둡니다.

 

Proposition. Define elementary symmetric polynomials of $n$ variables $\mathrm{\Lambda }=\{{\lambda }_1,\cdots ,{\lambda }_n\}$ by
$$\begin{array}{r}{c}_0=1,c_k=\sum _{\begin{array}{c}X\subset \mathrm{\Lambda }\\ |X|=k\end{array}}\prod _{\lambda \in X}\lambda (1\le k\le n)\end{array}$$
and similarly we define
$$\begin{array}{r}{s}_k=\sum _{j=1}^n{\lambda }_j^k(k\ge 0).\end{array}$$
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 $$\begin{array}{rl}(1-t{\lambda }_1)\cdots (1-t{\lambda }_n)& =\sum _{k=0}^{\mathrm{\infty }}(-1{)}^k{c}_k{t}^k.\end{array}$$ Similarly, we have $$\begin{array}{rl}\frac{1}{1-t{\lambda }_1}+\cdots +\frac{1}{1-t{\lambda }_n}& =\sum _{k=0}^{\mathrm{\infty }}{s}_k{t}^k.\end{array}$$ Now let $$\begin{array}{r}f(x)=(x-t{\lambda }_1)\cdots (x-t{\lambda }_n)=\sum _{k=0}^{\mathrm{\infty }}(-1{)}^k{c}_k{t}^k{x}^{n-k}.\end{array}$$ On the one hand, considering the log-derivative of $f(x)$, it is easy to confirm that $$\begin{array}{rl}{f}^{\prime }(1)& =(1-t{\lambda }_1)\cdots (1-t{\lambda }_n)(\frac{1}{1-t{\lambda }_1}+\cdots +\frac{1}{1-t{\lambda }_n})\\ & =\sum _{k=0}^{\mathrm{\infty }}(\sum _{j=0}^k(-1{)}^j{c}_j{s}_{k-j})t^k.\end{array}$$ On the other hand, $$\begin{array}{r}{f}^{\prime }(1)=\sum _{k=0}^{\mathrm{\infty }}(-1{)}^k(n-k)c_k{t}^k.\end{array}$$ Thus it follows that for any $k$ we have $$\begin{array}{r}\sum _{j=0}^k(-1{)}^j{c}_j{s}_{k-j}=(-1{)}^k(n-k)c_k.\end{array}$$ 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.