[Bhatia.Matrix Analysis.Solutions to Exercises and Problems]ExI.5.8

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.)

 

Solution. Let $$\bex A=\sex{\ba{cc} \al_1\\ \vdots\\ \al_n \ea},\quad B=\sex{\beta_1,\cdots,\beta_n}. \eex$$ Then $$\bex AB=\sex{\sef{\al_i,\beta_j}}. \eex$$ By Exercise I.5.7, $$\beex \bea |\per (AB)|^2 &=\sev{\per (\sef{\al_i,\beta_j})}^2\\ &\leq \per (\sef{\al_i,\al_j})\cdot \per (\sef{\beta_i,\beta_j})\\ &=\per(AA^*)\cdot \per(B^*B). \eea \eeex$$