Prove that for any vectors $$\bex u_1,\cdots,u_k,\quad v_1,\cdots,v_k, \eex$$ we have $$\bex |\det(\sef{u_i,v_j})|^2 \leq \det\sex{\sef{u_i,u_j}}\cdot \det \sex{\sef{v_i,v_j}}, \eex$$ $$\bex |\per(\sef{u_i,v_j})|^2 \leq \per\sex{\sef{u_i,u_j}}\cdot \per \sex{\sef{v_i,v_j}}. \eex$$
Solution. By Exercise I.5.1, $$\beex \bea |\det(\sef{u_i,v_j})|^2 &=\sev{ \sef{ u_1\wedge \cdots u_k,v_1\wedge \cdots \wedge v_k } }^2\\ &\leq \sen{ u_1\wedge \cdots \wedge u_k }^2\sen{ v_1\wedge \cdots \wedge v_k }^2\\ &=\det \sex{\sef{u_i,u_j}}\cdot \det \sex{\sef{v_i,v_j}}. \eea \eeex$$ Similarly, by Exercise I.5.5, we have $$\bex |\per(\sef{u_i,v_j})|^2 \leq \per\sex{\sef{u_i,u_j}}\cdot \per \sex{\sef{v_i,v_j}}. \eex$$
[Bhatia.Matrix Analysis.Solutions to Exercises and Problems]ExI.5.7
原文：http://www.cnblogs.com/zhangzujin/p/4112031.html