7. 设 $A_0\in M_n$ 正定, $A_i\in M_n$ 半正定, $i=1,\cdots,k$, 则 $$\bex \tr \sum_{j=1}^k \sex{\sum_{i=0}^jA_i}^{-2}A_j<\tr A_0^{-1}. \eex$$
证明: 记 $$\bex \sum_{i=0}^j A_i=B_j, \eex$$ 则 $$\beex \bea \tr\sex{\sum_{i=0}^j A_i}^{-2}A_j &=\tr \sex{B_j^{-2} (B_j-B_{j-1})}\\ &=\tr (B_j^{-1}-B_j^{-2}B_{j-1})\\ &=\sum_{i=1}^n s_j(B_j^{-1}) -\tr(B_j^{-2}B_{j-1})\\ &=\sum_{i=1}^n s_j(B_j^{-1}B_{j-1}^\frac{1}{2} \cdot B_{j-1}^{-\frac{1}{2}}) -\tr(B_j^{-2}B_{j-1})\\ &\leq\frac{1}{2}\sum_{i=1}^n s_i \sex{B_{j-1}^\frac{1}{2}B_j^{-2}B_{j-1}^\frac{1}{2} +B_{j-1}^{-1}} -\tr (B_j^{-2}B_{j-1})\quad\sex{\mbox{推论 4.18}}\\ &=\frac{1}{2}\tr \sex{B_{j-1}^\frac{1}{2}B_j^{-2}B_{j-1}^\frac{1}{2} +B_{j-1}^{-1}} -\tr (B_j^{-2}B_{j-1})\\ &=\frac{1}{2}\tr(B_j^{-2}B_{j-1}) +\frac{1}{2}\tr B_{j-1}^{-1} -\tr(B_j^{-2}B_{j-1})\quad\sex{\tr(XY)=\tr(YX)}\\ &=\frac{1}{2}\tr B_{j-1}^{-1} -\frac{1}{2} \tr(B_j^{-2}B_{j-1})\\ &=\frac{1}{2}\tr B_{j-1}^{-1} -\frac{1}{2} \tr(B_j^{-2}(B_j-A_j))\\ &=\frac{1}{2}\tr B_{j-1}^{-1} -\frac{1}{2} \tr B_j^{-1} +\frac{1}{2}\tr (B_j^{-2}A_j). \eea \eeex$$ 故 $$\beex \bea \tr B_j^{-2}A_j&=\tr B_{j-1}^{-1}-\tr B_j^{-1},\\ \tr \sum_{j=1}^k \sex{\sum_{i=0}^jA_i}^{-2}A_j &=\sum_{j=1}^k \tr B_j^{-2}A_j\\ &=\sum_{j=1}^k \sez{\tr B_{j-1}^{-1}-\tr B_j^{-1}}\\ &=\tr B_0^{-1} -\tr B_k^{-1}\\ &<\tr B_0^{-1}\\ &=\tr A_0^{-1}. \eea \eeex$$
