6. 设 $A,B\in M_n$ 半正定, 则 $$\bex s_j(A-B)\leq s_j\sex{ \sex{\ba{cc} A&0\\ 0&B \ea}},\quad j=1,\cdots,n. \eex$$

证明: $$\beex \bea s_j(A-B) &=\lm_j\sex{\sex{\ba{cc} A-B&0\\ 0&B-A \ea}}\quad\sex{\mbox{Hermite 矩阵的奇异值为其特征值的模长}}\\ &\leq \lm_j\sex{\sex{\ba{cc} A&0\\ 0&B \ea}}\quad\sex{\mbox{Weyl 单调性原理}}\\ &\quad\sex{\sex{\ba{cc} A&0\\ 0&B \ea}-\sex{\ba{cc} A-B&0\\ 0&B-A \ea}=\sex{\ba{cc} B&0\\ 0&A \ea}\geq 0}\\ &=s_j\sex{\sex{\ba{cc} A&0\\ 0&B \ea}}. \eea \eeex$$

[詹兴致矩阵论习题参考解答]习题4.6