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本帖最后由 hbghlyj 于 2022-10-26 17:52 编辑 来自wikipedia的Remark的第2条:
$c_i$是$n$阶矩阵$A$的特征值的初等对称多项式$e_{n-i}$.由牛顿恒等式,可以将$c_i$用$\operatorname{tr}(A^i)=λ_1^i+λ_2^i+⋯+λ_n^i$表示:
$$c_{i}=\sum _{k_{1},k_{2},\ldots ,k_{n}}\prod _{l=1}^{n}{\frac {(-1)^{k_{l}+1}}{l^{k_{l}}k_{l}!}}(\operatorname {tr} (A^{l}))^{k_{l}},$$其中,$k_l$为不同的非负整数,满足$\sum _{l=1}^{n}lk_{l}=n-i$.
例如,
$$\begin{aligned}c_1&=\operatorname{tr}A\\&=λ_1+λ_2+⋯+λ_n\end{aligned}\qquad(n≥2)$$
$$\begin{aligned}c_2&={\frac {1}{2}}\left((\operatorname {tr} A)^{2}-\operatorname {tr} (A^{2})\right)\\&={\frac {1}{2}}\left((λ_1+λ_2+⋯+λ_n)^{2}-(λ_1^2+λ_2^2+⋯+λ_n^2)\right)\\&=λ_1λ_2+λ_1λ_3+⋯+λ_{n-1}λ_n\end{aligned}\qquad(n≥3)$$
$$\begin{aligned}c_3&=\frac {1}{24}\!\left((\operatorname {tr} A)^{4}-6\operatorname {tr} (A^{2})(\operatorname {tr} A)^{2}+3(\operatorname {tr} (A^{2}))^{2}+8\operatorname {tr} (A^{3})\operatorname {tr} (A)-6\operatorname {tr} (A^{4})\right)\\&=\frac1{24}\Big(\left(λ_1+λ_2+⋯+λ_n\right)^4-6 \left(λ_1^2+λ_2^2+⋯+λ_n^2\right) \left(λ_1+λ_2+⋯+λ_n\right)^2+3 \left(λ_1^2+λ_2^2+⋯+λ_n^2\right)^2\\&\hphantom{=}+8 \left(λ_1^3+λ_2^3+⋯+λ_n^3\right) \left(λ_1+λ_2+⋯+λ_n\right)-6 \left(λ_1^4+λ_2^4+⋯+λ_n^4\right)\Big)\\&=λ_1λ_2λ_3+λ_1λ_2λ_4+⋯+λ_{n-2}λ_{n-1}λ_n\end{aligned}\qquad(n≥4)$$ |
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