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. $\\square$
随便看	DegreeMod2OfAMapping degree mod 2 of a mapping DegreeOfAlgebraicNumber degree of algebraic number DegreeOfAnAlgebraicNumber degree of an algebraic number DegreesOfFreedom degrees of freedom DehnsLemma DehnsTheorem DehnSurgery Dehn surgery Dehn’s lemma Dehn’s theorem DelayTheorem delay theorem DeletionOperationOnLanguages deletion operation on languages Delta1Bootstrapping DeltaDistribution delta distribution DeltaSystem delta system DeltaSystemLemma delta system lemma