proof of Schur’s inequality

By Schur’s theorem, a unitary matrix $U$ and an upper triangular matrix $T$ exist such that $A=UTU^{H}$ , $T$ being diagonal if and only if $A$ is normal.Then $A^{H}A=UT^{H}U^{H}UTU^{H}=UT^{H}TU^{H}$ , which means $A^{H}A$ and $T^{H}T$ are similar; so they have the same trace. We have:

$\\|A\\|_{F}^{2}=\\operatorname{Tr}(A^{H}A)=\\operatorname{Tr}(T^{H}T)=\\sum_{i=1}^{%n}\\left|\\lambda_{i}\\right|^{2}+\\sum_{i<j}\\left|t_{ij}\\right|^{2}=$

$=\\operatorname{Tr}(D^{H}D)+\\sum_{i<j}\\left|t_{ij}\\right|^{2}\\geq\\operatorname{%Tr}(D^{H}D)=\\|D\\|_{F}^{2}.$

If and only if $A$ is normal, $T=D$ and therefore equality holds. $\\square$

单词	ProofOfSchursInequality
释义	proof of Schur’s inequality By Schur’s theorem, a unitary matrix $U$ and an upper triangular matrix $T$ exist such that $A=UTU^{H}$ , $T$ being diagonal if and only if $A$ is normal.Then $A^{H}A=UT^{H}U^{H}UTU^{H}=UT^{H}TU^{H}$ , which means $A^{H}A$ and $T^{H}T$ are similar; so they have the same trace. We have: $\\\|A\\\|_{F}^{2}=\\operatorname{Tr}(A^{H}A)=\\operatorname{Tr}(T^{H}T)=\\sum_{i=1}^{%n}\\left\|\\lambda_{i}\\right\|^{2}+\\sum_{i<j}\\left\|t_{ij}\\right\|^{2}=$ $=\\operatorname{Tr}(D^{H}D)+\\sum_{i<j}\\left\|t_{ij}\\right\|^{2}\\geq\\operatorname{%Tr}(D^{H}D)=\\\|D\\\|_{F}^{2}.$ If and only if $A$ is normal, $T=D$ and therefore equality holds. $\\square$
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