中国科学院真题 第20176题

五、 $\left(20+5+5=30\right.$ 分）（1）$n$ 阶方阵 $A$ 能表成：$A=H+K$ ，其中 $H=\bar{H}^{T}, K=\bar{K}^{T}$ ，矩阵 $\bar{B}^{T}$ 表示矩阵 $B$ 的共轭转置。设 $a, h, k$ 分别是 $A, H, K$ 中元素最大模，若 $z= x+i y(x, y \in \mathbb{R})$ 是 $A$ 的任意特征值．求证： $$ |z| \leq n a,|x| \leq n h,|y| \leq n k $$ （2）求证：Hermite矩阵的特征值都是实数； （3）求证：反对称矩阵的非零特征值都是纯虚数． 证明：（1）设 $A=\left(a_{i j}\right) \in \mathbb{C}^{n \times n}, A \xi=z \xi$ 且 $\xi=\left(\varepsilon_{1}, \varepsilon_{2}, \cdots, \varepsilon_{n}\right)^{T} \neq 0,\left|\varepsilon_{k}\right|=\max \left\{\varepsilon_{1}\right.$ ， $\left.\varepsilon_{2}, \cdots, \varepsilon_{n}\right\}$ ，则 $$ z \varepsilon_{k}=a_{k 1} \varepsilon_{1}+a_{k 2} \varepsilon_{2}+\cdots+a_{k n} \varepsilon_{n} $$ 所以 $$ |z| \cdot\left|\varepsilon_{k}\right|=\left|z \varepsilon_{k}\right|=\left|a_{k 1} \varepsilon_{1}+a_{k 2} \varepsilon_{2}+\cdots+a_{k n} \varepsilon_{n}\right| $$ $$ \begin{aligned} & \quad \leq\left|a_{k 1}\right| \cdot\left|\varepsilon_{1}\right|+\left|a_{k 2}\right| \cdot\left|\varepsilon_{2}\right|+\cdots+\left|a_{k n}\right| \cdot\left|\varepsilon_{n}\right| \leq\left|a_{k 1}\right| \cdot\left|\varepsilon_{k}\right|+\left|a_{k 2}\right| \cdot\left|\varepsilon_{k}\right|+\cdots+\left|a_{k n}\right| \cdot\left|\varepsilon_{k}\right| \\ & \because \xi \neq 0 \quad \therefore\left|\varepsilon_{k}\right|>0 \quad \therefore|z| \leq\left|a_{k 1}\right|+\left|a_{k 2}\right|+\cdots+\left|a_{k n}\right| \leq n a \\ & \because A=H+K \therefore H \xi+K \xi=A \xi=(x+i y) \xi=x \xi+i y \xi \\ & \therefore H \xi-x \xi=-K \xi+i y \xi \quad \therefore \bar{\xi}^{T}\left(H-x I_{n}\right) \xi=\bar{\xi}^{T}\left(-K+i y I_{n}\right) \xi \\ & \because \bar{K}^{T}=-K, \bar{H}^{T}=H \\ & \therefore \bar{\xi}^{T}\left(H-x I_{n}\right) \xi=\bar{\xi}^{T}\left(-K+i y I_{n}\right) \xi \\ & \quad \bar{\xi}^{T}\left(H-x I_{n}\right) \xi=\bar{\xi}^{T}\left(K-i y I_{n}\right) \xi=-\bar{\xi}^{T}\left(-K+i y I_{n}\right) \xi=-\bar{\xi}^{T}\left(H-x I_{n}\right) \xi \\ & \therefore \bar{\xi}^{T}\left(H-x I_{n}\right) \xi=\bar{\xi}^{T}\left(-K+i y I_{n}\right) \xi=0 \end{aligned} $$ 设 $H=\left(h_{i j}\right), K=\left(k_{i j}\right) \in \mathbb{C}^{n \times n}$ ，则由上式可得 $$ \begin{aligned} & h_{k 1} \varepsilon_{1}^{2}+h_{k 2} \varepsilon_{2}^{2}+\cdots+\left(h_{k k}-x\right) \varepsilon_{k}^{2}+\cdots+h_{k n} \varepsilon_{n}^{2}=0 \\ & k_{k 1} \varepsilon_{1}^{2}+k_{k 2} \varepsilon_{2}^{2}+\cdots+\left(k_{k k}-i y\right) \varepsilon_{k}^{2}+\cdots+k_{k n} \varepsilon_{n}^{2}=0 \end{aligned} $$ 所以 $$ \begin{aligned} x \varepsilon_{k}^{2} & =h_{k 1} \varepsilon_{1}^{2}+h_{k 2} \varepsilon_{2}^{2}+\cdots+h_{k n} \varepsilon_{n}^{2} \\ i y \varepsilon_{k}^{2} & =k_{k 1} \varepsilon_{1}^{2}+k_{k 2} \varepsilon_{2}^{2}+\cdots+k_{k n} \varepsilon_{n}^{2} \end{aligned} $$ 故 $$ \begin{aligned} |x| \cdot\left|\varepsilon_{k}\right|^{2} & =\left|x \varepsilon_{k}^{2}\right|=\left|h_{k 1} \varepsilon_{1}^{2}+h_{k 2} \varepsilon_{2}^{2}+\cdots+h_{k n} \varepsilon_{n}^{2}\right| \\ & \leq\left|h_{k 1} \varepsilon_{1}^{2}\right|+\left|h_{k 2} \varepsilon_{2}^{2}\right|+\cdots+\left|h_{k n} \varepsilon_{n}^{2}\right|=\left|h_{k 1}\right| \cdot\left|\varepsilon_{1}^{2}\right|+\left|h_{k 2}\right| \cdot\left|\varepsilon_{2}^{2}\right|+\cdots+\left|h_{k n}\right| \cdot\left|\varepsilon_{n}^{2}\right| \\ & \leq\left|h_{k 1}\right| \cdot\left|\varepsilon_{k}^{2}\right|+\left|h_{k 2}\right| \cdot\left|\varepsilon_{k}^{2}\right|+\cdots+\left|h_{k n}\right| \cdot\left|\varepsilon_{k}^{2}\right| \leq n h\left|\varepsilon_{k}^{2}\right| \\ |y| \cdot\left|\varepsilon_{k}\right|^{2} & =\left|i y \varepsilon_{k}^{2}\right|=\left|k_{k 1} \varepsilon_{1}^{2}+k_{k 2} \varepsilon_{2}^{2}+\cdots+k_{k n} \varepsilon_{n}^{2}\right| \\ & \leq\left|h_{k 1} \varepsilon_{1}^{2}\right|+\left|k_{k 2} \varepsilon_{2}^{2}\right|+\cdots+\left|k_{k n} \varepsilon_{n}^{2}\right|=\left|k_{k 1}\right| \cdot\left|\varepsilon_{1}^{2}\right|+\left|k_{k 2}\right| \cdot\left|\varepsilon_{2}^{2}\right|+\cdots+\left|k_{k n}\right| \cdot\left|\varepsilon_{n}^{2}\right| \\ & \leq\left|k_{k 1}\right| \cdot\left|\varepsilon_{k}^{2}\right|+\left|k_{k 2}\right| \cdot\left|\varepsilon_{k}^{2}\right|+\cdots+\left|k_{k n}\right| \cdot\left|\varepsilon_{k}^{2}\right| \leq n k\left|\varepsilon_{k}^{2}\right| \end{aligned} $$ 所以，$|x| \leq n h,|y| \leq n k$ 。 （2）设 $\lambda$ 是任意 Hermite 矩阵 $B$ 的任一特征值，并设 $B \xi=\lambda \xi$ 且 $\xi \neq 0$ 。则 $$ \lambda \bar{\xi}^{T} \xi=\bar{\xi}^{T} B \xi=\bar{\xi}^{T} \bar{B}^{T} \xi={\overline{\bar{\xi}^{T}} B \xi}^{T}=\overline{\lambda \xi}^{T} \xi $$ $\because \xi \neq 0 \quad \therefore \bar{\xi} \xi>0 \quad \therefore \lambda=\bar{\lambda} \quad \therefore \lambda \in \mathbb{R}$ ，由 $\lambda$ 的任意性可得 Hermite 矩阵的特征值都是实数。 （3）设 $\lambda$ 是任意反对称矩阵 $C$ 的任一非零特征值（如果有的话），并设 $B \xi=\lambda \xi$ 且 $\xi \neq 0$ ．则 $$ \lambda \bar{\xi}^{T} \xi=\bar{\xi}^{T} C \xi=-\bar{\xi}^{T} \bar{C}^{T} \xi=-{\overline{\xi^{T}} C \xi}^{T}=-\bar{\lambda} \bar{\xi}^{T} \xi $$ $\because \xi \neq 0 \quad \therefore \bar{\xi} \xi>0 \quad \therefore \lambda=-\bar{\lambda} \quad \therefore \lambda$ 是纯虚数，由 $\lambda$ 的任意性可得 反对称矩阵的非零特征值都是纯虚数。