张祖锦常用结论09可交换的复方阵可以同时相似上三角化

zhangzujin

手机查看请在浏览器中打开, 到了支付页面请截图, 并用支付宝或微信扫描之, 稍等后获得金钱, 即可购买. 偶偶因为网络问题充值不成功, 请与微信 pdezhang 联系, 发送论坛昵称与付款时间即可处理, 稍安勿躁. 购买后刷新网页才能正常显示数学公式.

张祖锦常用结论可交换的复方阵可以同时相似上三角化

设 $\displaystyle A,B$ 都是 $\displaystyle n$ 阶复矩阵. 证明: 若 $\displaystyle A$ 与 $\displaystyle B$ 乘积可交换, 那么存在 $\displaystyle n$ 阶可逆矩阵 $\displaystyle P$ 使得 $\displaystyle P^{-1}AP$ 和 $\displaystyle P^{-1}BP$ 都是上三角阵.

纸质资料/答疑/pdf1/pdf2 / 对 $\displaystyle n$ 作数学归纳法. 当 $\displaystyle n=1$ 时, 结论自明. 假设对阶数 $\displaystyle \leq n-1$ 的复矩阵结论成立, 则对 $\displaystyle n$ 阶复矩阵 $\displaystyle A,B$, 设 $\displaystyle \mathbb{C}^n$ 上的线性变换 $\displaystyle \mathscr{A}, \mathscr{B}$ 在基 $\displaystyle e_1,\cdots,e_n$ 下的矩阵分别为 $\displaystyle A,B$, 则 $\displaystyle \mathscr{A}\mathscr{B}=\mathscr{B}\mathscr{A}$. 任取 $\displaystyle \mathscr{A}$ 的特征值 $\displaystyle \lambda\in\mathbb{C}$, 令

$$\begin{aligned} V_\lambda=\left\{\alpha\in\mathbb{C}^n; \mathscr{A}\alpha=\lambda\alpha\right\}, \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$

则由

$$\begin{aligned} \alpha\in V_\lambda&\Rightarrow \mathscr{A}\mathscr{B}\alpha=\mathscr{B}\mathscr{A}\alpha=\mathscr{B}(\lambda\alpha) =\lambda \mathscr{B}\alpha\\\\ &\Rightarrow \mathscr{B}\alpha\in V_\lambda \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$

知 $\displaystyle V_\lambda$ 上的线性变换 $\displaystyle \mathscr{B}|_{V_\lambda}$ 也有特征值 $\displaystyle \mu$ 及相应的特征向量 $\displaystyle \eta_1$, 将 $\displaystyle \eta_1$ 扩充为 $\displaystyle \mathbb{C}^n$ 的一组基 $\displaystyle \eta_1,\cdots,\eta_n$, 则

$$\begin{aligned} \mathscr{A}(\eta_1,\cdots,\eta_n)=&(\eta_1,\cdots,\eta_n)\left(\begin{array}{cccccccccc}\lambda&\star \\\\ 0&A_1\end{array}\right),\\\\ \mathscr{B}(\eta_1,\cdots,\eta_n)=&(\eta_1,\cdots,\eta_n)\left(\begin{array}{cccccccccc}\mu&\star \\\\ 0&B_1\end{array}\right). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$

又由

$$\begin{aligned} \mathscr{A}(\eta_1,\cdots,\eta_n)&=\mathscr{A}(e_1,\cdots,e_n)(\eta_1,\cdots\eta_n)\\\\ &=(e_1,\cdots,e_n)A(\eta_1,\cdots,\eta_n)\\\\ &\stackrel{P_1=(\eta_1,\cdots,\eta_n)}{=} (\eta_1,\cdots,\eta_n)P_1^{-1}AP_1. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$

于是

$$\begin{aligned} P_1^{-1}AP_1=\left(\begin{array}{cccccccccc}\lambda&\star \\\\ 0&A_1\end{array}\right), P_1^{-1}BP_1=\left(\begin{array}{cccccccccc}\mu&\star \\\\ 0&B_1\end{array}\right). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$

再由 $\displaystyle AB=BA$ 知 $\displaystyle A_1B_1=B_1A_1$. 而由归纳假设, 存在可逆阵 $\displaystyle P_2$, 使得

$$\begin{aligned} P_2^{-1}A_1P_2=U_1, P_2^{-1}B_1P_2=U_2 \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$

均为上三角阵. 令

$$\begin{aligned} P=P_1\left(\begin{array}{cccccccccc}1&\\\\ &P_2\end{array}\right), \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$

则 $\displaystyle P$ 可逆, 且

$$\begin{aligned} P^{-1}AP=\left(\begin{array}{cccccccccc}\lambda&\star \\\\ &U_1\end{array}\right), P^{-1}BP=\left(\begin{array}{cccccccccc}\mu&\star \\\\ &U_2\end{array}\right) \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$

都是上三角阵. 跟锦数学微信公众号. 在线资料/公众号/