## 利用牛頓恆等式聯繫特徵多項式係數與冪矩陣跡數

$A$ 為一 $n\times n$ 階矩陣。定義 $A$ 的特徵多項式為

$\displaystyle p(t)=\det(tI-A)=t^n+a_1t^{n-1}+\cdots+a_{n-1}t+a_n=\sum_{k=0}^na_kt^{n-k}$

\begin{aligned} p(t)&=(t-\lambda_1)(t-\lambda_2)(t-\lambda_3)\\ &=t^3-(\lambda_1+\lambda_2+\lambda_3)t^2+(\lambda_1\lambda_2+\lambda_1\lambda_3+\lambda_2\lambda_3)t-\lambda_1\lambda_2\lambda_3. \end{aligned}

\begin{aligned} a_1&=-(\lambda_1+\lambda_2+\lambda_3)=-s_1(\lambda_1,\lambda_2,\lambda_3)\\ a_2&=\lambda_1\lambda_2+\lambda_1\lambda_3+\lambda_2\lambda_3=s_2(\lambda_1,\lambda_2,\lambda_3)\\ a_3&=-\lambda_1\lambda_2\lambda_3=-s_3(\lambda_1,\lambda_2,\lambda_3). \end{aligned}

\begin{aligned} t_1&=\lambda_1+\lambda_2+\lambda_3=-a_1\\ t_2&=\lambda_1^2+\lambda_2^2+\lambda_3^2=(\lambda_1+\lambda_2+\lambda_3)^2-2(\lambda_1\lambda_2+\lambda_1\lambda_3+\lambda_2\lambda_3)=-t_1a_1-2a_2\\ t_3&=\lambda_1^3+\lambda_2^3+\lambda_3^3=(\lambda_1+\lambda_2+\lambda_3)(\lambda_1^2+\lambda_2^2+\lambda_3^2-\lambda_1\lambda_2-\lambda_1\lambda_3-\lambda_2\lambda_3)+3\lambda_1\lambda_2\lambda_3\\ &=-t_2a_1-t_1a_2-3a_3, \end{aligned}

\begin{aligned} a_1&=-t_1\\ 2a_2&=-t_1a_1-t_2\\ 3a_3&=-t_2a_1-t_1a_2-t_3. \end{aligned}

\displaystyle\begin{aligned} ka_k&=-\sum_{j=1}^kt_j(\lambda_1,\ldots,\lambda_n)a_{k-j},~~~k=1,\ldots,n,\\ 0&=\sum_{j=k-n}^kt_j(\lambda_1,\ldots,\lambda_n)a_{k-j},~~~k>n.\end{aligned}

$\displaystyle p(t)=\sum_{k=0}^na_kt^{n-k}=\prod_{i=1}^n(t-\lambda_i)$

$t=1/x$ 代入上式，等號兩邊同乘以 $x^n$，則有

$\displaystyle \sum_{k=0}^na_kx^k=\prod_{i=1}^n\left(1-\lambda_ix\right)$

\displaystyle\begin{aligned} \sum_{k=0}^nka_kx^k&=x\sum_{i=1}^n\left((-\lambda_i)\prod_{j\neq i}\left(1-\lambda_jx\right)\right)\\ &=-\left(\sum_{i=1}^n\frac{\lambda_ix}{1-\lambda_ix}\right)\prod_{j=1}^n\left(1-\lambda_jx\right)\\ &=-\left(\sum_{i=1}^n\sum_{j=1}^{\infty}(\lambda_ix)^j\right)\left(\sum_{l=0}^na_lx^l\right)\\ &=-\left(\sum_{j=1}^{\infty}\left(\sum_{i=1}^n\lambda_i^j\right)x^j\right)\left(\sum_{l=0}^na_lx^l\right)\\ &=-\left(\sum_{j=1}^{\infty}t_j(\lambda_1,\ldots,\lambda_n)x^j\right)\left(\sum_{l=0}^na_lx^l\right). \end{aligned}

\displaystyle\begin{aligned} ka_k&=-\sum_{j=1}^kt_j(\lambda_1,\ldots,\lambda_n)a_{k-j},~~~k=1,\ldots,n,\\ 0&=\sum_{j=k-n}^kt_j(\lambda_1,\ldots,\lambda_n)a_{k-j},~~~k>n,\end{aligned}

\begin{aligned} a_1&=-t_1\\ t_1a_1+2a_2&=-t_2\\ t_2a_1+t_1a_2+3a_3&=-t_3\\ &\vdots\\ t_{n-1}a_1+t_{n-2}a_2+\cdots+t_1a_{n-1}+na_n&=-t_n. \end{aligned}

$\begin{bmatrix} 1&0&0&\cdots&0&0\\ t_1&2&0&\cdots&0&0\\ t_2&t_1&3&\cdots&0&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ t_{k-2}&t_{k-3}&t_{k-4}&\cdots&k-1&0\\ t_{k-1}&t_{k-2}&t_{k-3}&\cdots&t_1&k \end{bmatrix}\begin{bmatrix} a_1\\ a_2\\ a_3\\ \vdots\\ a_{k-1}\\ a_k \end{bmatrix}=\begin{bmatrix} -t_1\\ -t_2\\ -t_3\\ \vdots\\ -t_{k-1}\\ -t_k \end{bmatrix}$

$\displaystyle a_k=\frac{\begin{vmatrix} 1&0&\cdots&0&-t_1\\ t_1&2&\cdots&0&-t_2\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ t_{k-2}&t_{k-3}&\cdots&k-1&-t_{k-1}\\ t_{k-1}&t_{k-2}&\cdots&t_1&-t_k \end{vmatrix}}{\begin{vmatrix} 1&0&\cdots&0&0\\ t_1&2&\cdots&0&0\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ t_{k-2}&t_{k-3}&\cdots&k-1&0\\ t_{k-1}&t_{k-2}&\cdots&t_1&k \end{vmatrix}}=\frac{1}{k!}\begin{vmatrix} 1&0&\cdots&0&-t_1\\ t_1&2&\cdots&0&-t_2\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ t_{k-2}&t_{k-3}&\cdots&k-1&-t_{k-1}\\ t_{k-1}&t_{k-2}&\cdots&t_1&-t_k \end{vmatrix}$

$\displaystyle a_k=\frac{(-1)^k}{k!}\begin{vmatrix} t_1&1&0&\cdots&0\\ t_2&t_1&2&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ t_{k-1}&t_{k-2}&t_{k-3}&\cdots&k-1\\ t_k&t_{k-1}&t_{k-2}&\cdots&t_1 \end{vmatrix},~~~k=1,\ldots,n$

$\displaystyle a_1=-t_1,~~a_2=\frac{1}{2!}\begin{vmatrix} t_1&1\\ t_2&t_1 \end{vmatrix}=\frac{1}{2}\left(t_1^2-t_2\right),~~a_3=-\frac{1}{3!}\begin{vmatrix} t_1&1&0\\ t_2&t_1&2\\ t_3&t_2&t_1 \end{vmatrix}=\frac{1}{6}\left(-t_1^3+3t_1t_2-2t_3\right)$

$k=n$，則 $a_n=(-1)^n\prod_{i=1}^n\lambda_i=(-1)^n\det A$，即 $\det A=(-1)^na_n$。所以，$\det A$ 可表示成 $t_j=\mathrm{trace}(A^j)$ 構成的行列式：

$\displaystyle \det A=\frac{1}{n!}\begin{vmatrix} t_1&1&0&\cdots&0\\ t_2&t_1&2&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ t_{n-1}&t_{n-2}&t_{n-3}&\cdots&n-1\\ t_n&t_{n-1}&t_{n-2}&\cdots&t_1 \end{vmatrix}$

[1] 維基百科：Newton’s identities
[2] Dan Kalman, A matrix proof of Newton’s identities, Mathematics Magazine, 73, pp. 313-315, 2000.
[3] 維基百科：Hessenberg matrix

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