有关矩阵级数的一道高代考研题试解

此题系中山大学2017年高代考研题,由网友水清浅向我提问…个人试解如下
$Problem$:
$设A是一个n阶实矩阵,其有n个绝对值小于1的实特征值,证明:$
$$ln(det(I-A)=-\sum _{k=1}^{\infty }\frac{1}{k}tr(A^k)$$
$Proof$:
\qquad\quad$首先证明如下引理$:
\qquad\quad$任意n阶方阵A必相似于上三角阵T,其主对角线元素是A的n个特征$
\qquad\quad$值:$${\lambda }_1,\cdots ,{\lambda }_n$,$即存在n阶可逆方阵P使得$
$$P^{-1}AP=\left(\begin{array}{cccc}{\lambda }_1& \ast & \ast & \ast \\ 0& {\lambda }_2& \cdots & \ast \\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & {\lambda }_n\end{array}\right)$$
\qquad\quad$证明如下:$
\qquad\quad$只需要证明对于任意n阶方阵A,存在n阶可逆方阵P和n$$阶上三角方$\quad$阵T,使得A=P^{-1}TP.$$采用数学归纳法.$
$1^0$,$当n=1时,结论显然成立.$
$2^0$,$设对于n-1阶方阵结论成立,$$A是任一n阶方阵,设v是A的一个特征向量$,$v̸=0$,$Av=\lambda v.$$取{\mathbb{R}}^n$$的一组基,以v为第一个基向量$,$于是有$
$$S^{-1}AS=\left(\begin{array}{cccc}\lambda & \ast & \cdots & \ast \\ 0& & & \\ ⋮& & A_0& \\ 0& & & \end{array}\right)$$
$其中S是某个n阶可逆方阵.依归纳假设,对于n-1阶方阵A_0$,$存在某个$$n-1$$阶可逆方阵S_0使得S_0^{-1}{A}_0{S}_0=T_0$.$令$
$$S_1=\left(\begin{array}{cccc}1& 0& \cdots & 0\\ 0& & & \\ ⋮& & S_0& \\ 0& & & \end{array}\right)$$
$因而$
$$S_1^{-1}=\left(\begin{array}{cccc}1& 0& \cdots & 0\\ 0& & & \\ ⋮& & S_0^{-1}& \\ 0& & & \end{array}\right)$$
$故$
$$\begin{gathered} S_1^{-1}\left(S^{-1}AS\right)S_1=S_1^{-1}\left(\begin{array}{cccc}\lambda & \ast & \cdots & \ast \\ 0& & & \\ ⋮& & A_0& \\ 0& & & \end{array}\right)S_1 \\ =\left(\begin{array}{cccc}\lambda & \ast & \cdots & \ast \\ 0& & & \\ ⋮& & T_0& \\ 0& & & \end{array}\right) \end{gathered}$$
\qquad\quad$证明完毕.$

\qquad\quad$下面证明本题:$
$$\begin{gathered} 考虑T=\left(\begin{array}{cccc}{\lambda }_1& \ast & \cdots & \ast \\ 0& {\lambda }_2& \cdots & \ast \\ ⋮& ⋮& \ddots & ⋮\\ 0& \cdots & \cdots & {\lambda }_n\end{array}\right) \\ 则T^k=\left(\begin{array}{cccc}{\lambda }_1^k& \ast & \cdots & \ast \\ 0& {\lambda }_2^k& \cdots & \ast \\ ⋮& ⋮& \ddots & ⋮\\ 0& \cdots & \cdots & {\lambda }_n^k\end{array}\right) \\ 从而\sum _{k=0}^{\infty }\frac{T^k}{k!}=\left(\begin{array}{cccc}{e}^{{\lambda }_1}& \ast & \cdots & \ast \\ 0& e^{{\lambda }_2}& \cdots & \ast \\ ⋮& ⋮& \ddots & ⋮\\ 0& \cdots & \cdots & e^{{\lambda }_n}\end{array}\right) \\ \det \left(\sum _{k=0}^{\infty }\frac{T^k}{k!}\right)=e^{{\lambda }_1+{\lambda }_2+\cdots {\lambda }_n} \\ \det \left(e^T\right)=e^{tr\left(T\right)} \end{gathered}$$
\qquad\quad$由引理及上述讨论知$
$$e^A=e^{P^{-1}TP}=\sum _{k=0}^{\infty }\frac{(P^{-1}TP{)}^k}{k!}=P^{-1}{e}^{T}P$$\qquad\quad$因而$
$$det(e^A)=det(P^{-1}{e}^{T}P)=det(e^T)=e^{tr(T)}=e^{tr(A)}$$\qquad\quad$故$
$$\begin{gathered} det\left(I-A\right)=det\left(e^{ln\left(I-A\right)}\right)=det\left(\exp \left(-\sum _{k=1}^{\infty }\frac{A^k}{k}\right)\right) \\ =\prod _{k=1}^{\infty }\det \left(-\exp \left(\frac{A^k}{k}\right)\right) \\ =\prod _{k=1}^{\infty }\exp \left(-tr\frac{A^k}{k}\right) \\ =\exp \left(\sum -\frac{tr{A}^k}{k}\right) \end{gathered}$$