[詹兴致矩阵论习题参考解答]习题5.4

4. (G.M. Krause) 令 $$\bex \lm_1=1,\quad \lm_2=\frac{4+5\sqrt{3}I}{13},\quad \lm_3=\frac{-1+2\sqrt{3}i}{13},\quad v=\sex{\sqrt{\frac{5}{8}},\frac{1}{2},\sqrt{\frac{1}{8}}}^T. \eex$$ 再令 $$\bex A=\diag(\lm_1,\lm_2,\lm_3),\quad U=I-2vv^T,\quad B=-U^*AU, \eex$$ 则 $U$ 为酉矩阵, $A,B$ 为正规矩阵. 验证 $$\bex \rd (\sigma(A),\sigma(B))=\sqrt{\frac{28}{13}},\quad \sen{A-B}_\infty =\sqrt{\frac{27}{13}}. \eex$$ 于是, 对于这一对正规矩阵 $A,B$, $$\bex \rd (\sigma(A)),\sigma(B))>\sen{A-B}_\infty. \eex$$

 

 

证明: 记 $$\bex \sigma(A)=\sed{\al_1,\al_2,\al_3} =\sed{\lm_1,\lm_2,\lm_3},\quad \sigma(B)=\sed{\beta_1,\beta_2,\beta_3} =\sed{-\lm_1,-\lm_2,-\lm_3}. \eex$$ 则 $$\beex \bea \sigma=\sed{1,2,3}&\ra \max_i|\al_i-\beta_{\sigma(i)}| =2,\\ \sigma=\sed{1,3,2}&\ra \max_i|\al_i-\beta_{\sigma(i)}| =2,\\ \sigma=\sed{2,1,3}&\ra \max_i|\al_i-\beta_{\sigma(i)}| =\sqrt{\frac{28}{13}},\\ \sigma=\sed{2,3,1}&\ra \max_i|\al_i-\beta_{\sigma(i)}| =\sqrt{\frac{28}{13}},\\ \sigma=\sed{3,1,2}&\ra \max_i|\al_i-\beta_{\sigma(i)}| =\sqrt{\frac{28}{13}},\\ \sigma=\sed{3,2,1}&\ra \max_i|\al_i-\beta_{\sigma(i)}| =\sqrt{\frac{28}{13}}. \eea \eeex$$ 故 $$\bex \rd(\sigma(A),\sigma(B))=\sqrt{\frac{28}{13}}. \eex$$ 又通过数学软件可以算出 $(A-B)^*(A-B)$ 的特征值为 $$\bex \frac{27}{13},\quad \frac{27}{13},\quad \frac{4}{13}, \eex$$ 而 $$\bex \sen{A-B}_\infty=\sqrt{\frac{27}{13}}. \eex$$