在 Rajendra Bhatia 的 Matrix Analysis 中, Exercise I.5.8 说: Prove that for any matrices $A,B$ we have $$ex |per (AB)|^2leq per (AA^*)cdot per (B^*B). eex$$ (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: $$eex ea |per (AB)|^2 &=sev{sum_sigma c_{1sigma(1)}cdot c_{nsigma(n)}}^2quadsex{C=AB}\ &=sev{ sum_{sigma} sez{sum_{k_1} a_{1k_1}b_{k_1sigma(1)} cdots sum_{k_n} a_{nk_n}b_{k_nsigma(n)} }}^2\ &=sev{ sum_{k_1cdots k_n}sez{ (a_{1k_1}cdots a_{nk_n})cdot sex{sum_{sigma} b_{k_1sigma(1)}cdots b_{k_nsigma(n)}} }}^2\ &leq sum_{k_1cdots k_n} |a_{1k_1}cdots a_{nk_n}|^2 cdot sum_{k_1cdots k_n}sev{ sum_sigma | b_{k_1sigma(1)}cdots b_{k_nsigma(n)} }^2\ &=sum_{k_1}|a_{1k_1}|^2 cdots sum_{k_n}|a_{nk_n}|^2 cdot sum_{k_1cdots k_n} sex{sum_sigma ar b_{k_1sigma(1)}cdots ar b_{k_nsigma(n)} cdot sum_ au b_{k_1 au(1)}cdots b_{k_n au(n)}}\ &= ilde a_{11}cdots ilde a_{nn} sum_{sigma, au} sex{ sum_{k_1} ar b_{k_1sigma(1)}b_{k_1 au(1)} cdots sum_{k_n}ar b_{k_nsigma(n)}b_{k_n au(n)} }quadsex{AA^*= ilde A}\ &= ilde a_{11}cdots ilde a_{nn} sum_{sigma}sum_{ au} ilde b_{sigma(1) au(1)} cdots ilde b_{sigma(n) au(n)}quadsex{B^*B= ilde B}\ &=n!cdot ilde a_{11}cdots ilde a_{nn}cdot per(B^*B). eea eeex$$