[Bhatia.Matrix Analysis.Solutions to Exercises and Problems]ExI.5.9

(Schur's Theorem) If $A$ is positive, then $$\bex \per(A)\geq \det A. \eex$$

 

Solution. By Exercise I.2.2, $A=T^*T$ for some upper triangular $T$ with non-negative diagonals. Thus $$\beex \bea \det A&=\det T^*\cdot \det T\\ &=\per T^*\cdot \per T\\ &=\per(T^*I)\cdot \per(I\cdot T)\\ &\leq \sqrt{\per(T^*T)\cdot \per (I^*I)}\cdot \sqrt{\per(II^*)\cdot \per (T^*T)}\\ &=\per(T^*T)\\ &=\per(A). \eea \eeex$$