在 Rajendra Bhatia 的 Matrix Analysis 中, Exercise I.5.8 说: Prove that for any matrices $A,B$ we have $$\bex |\per (AB)|^2\leq \per (AA^*)\cdot \per (B^*B). \eex$$ (The corresponding relation for determinants is an easy equality.)
到目前为止, 我还没能证出. 不过得到一个貌似更弱的结论. 请看证明: $$\beex \bea |\per (AB)|^2 &=\sev{\sum_\sigma c_{1\sigma(1)}\cdot c_{n\sigma(n)}}^2\quad\sex{C=AB}\\ &=\sev{ \sum_{\sigma} \sez{\sum_{k_1} a_{1k_1}b_{k_1\sigma(1)} \cdots \sum_{k_n} a_{nk_n}b_{k_n\sigma(n)} }}^2\\ &=\sev{ \sum_{k_1\cdots k_n}\sez{ (a_{1k_1}\cdots a_{nk_n})\cdot \sex{\sum_{\sigma} b_{k_1\sigma(1)}\cdots b_{k_n\sigma(n)}} }}^2\\ &\leq \sum_{k_1\cdots k_n} |a_{1k_1}\cdots a_{nk_n}|^2 \cdot \sum_{k_1\cdots k_n}\sev{ \sum_\sigma | b_{k_1\sigma(1)}\cdots b_{k_n\sigma(n)} }^2\\ &=\sum_{k_1}|a_{1k_1}|^2 \cdots \sum_{k_n}|a_{nk_n}|^2 \cdot \sum_{k_1\cdots k_n} \sex{\sum_\sigma \bar b_{k_1\sigma(1)}\cdots \bar b_{k_n\sigma(n)} \cdot \sum_\tau b_{k_1\tau(1)}\cdots b_{k_n\tau(n)}}\\ &=\tilde a_{11}\cdots \tilde a_{nn} \sum_{\sigma,\tau} \sex{ \sum_{k_1} \bar b_{k_1\sigma(1)}b_{k_1\tau(1)} \cdots \sum_{k_n}\bar b_{k_n\sigma(n)}b_{k_n\tau(n)} }\quad\sex{AA^*=\tilde A}\\ &=\tilde a_{11}\cdots \tilde a_{nn} \sum_{\sigma}\sum_{\tau} \tilde b_{\sigma(1)\tau(1)} \cdots \tilde b_{\sigma(n)\tau(n)}\quad\sex{B^*B=\tilde B}\\ &=n!\cdot \tilde a_{11}\cdots \tilde a_{nn}\cdot \per(B^*B). \eea \eeex$$
