满秩方阵\(A\)的对数\(\log A\)

elim

题 1： 设\(\,A= I + K,\;\; K = \lambda\big(\delta(j-i-1)\big)_{n\times n},\;\delta(x)=\small\begin{cases}0,& x\ne 0,\\ 1,& x=0.\end{cases}\)
\(\qquad\) 试证\(\; A = e^B,\;\;B=\small\displaystyle\sum_{m=1}^{n-1}\frac{(-1)^{m-1}}{m}K^m\)

elim

风花飘飘 发表于 2021-12-17 11:59
是否可逆？

A 必须是可逆的， log A 未必可逆。

elim

题： 给定方阵\(\,A,\,\)定义\(\,\log A\,\)为方程\(\,A=e^X\) 的解.
\(\quad\)给出方程有解的条件及通解公式.
解：\(\because\;e^X\,\)恒可逆，\(\therefore\) 仅当\(\,A\)可逆时方程有解.
\(\quad\)首先考虑\(\small\,G=I+K,\,I=(\delta(i-j))_{n\times n},\,K=\eta(\delta(j-i-1))_{n\times n}\)
\(\quad\small\delta(x)=\begin{cases}1,& x=0,\\ 0,& x\ne 0.\end{cases}\) 易见\(\;K^m=\eta^m(\delta(j-i-m))_{n\times n},\)
\(\quad\small K^n=\mathbf{0}.\;\)令\(\small\;E=\displaystyle\sum_{m=1}^{n-1} {\scriptsize\frac{(-1)^{m-1}}{m}}K^m\) 故据(*)，\(e^E\small=I+K=G\)
\(\quad\)故对Jordan块有\(\small\,J=\lambda I(I+K)=e^{(\ln\lambda)I+E}=e^{\log J}\;(\eta=\lambda^{-1}).\)
\(\quad\)进而对一般的可逆阵\(\,A=T^{-1}\text{diag}(J_1,\ldots,J_m)T\) 有
\(\quad \log A = T^{-1}\text{diag}(\log J_1,\ldots,\log J_m)T.\) (\(J_j\) 是Jordan块)

elim

显然 4 楼给出了主贴的解. 现在的问题是,  求所有\(\Delta\)使 \(e^{\Delta+\log A }=A,\; \)