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. $\square$

Title	proof of Schur’s inequality
Canonical name	ProofOfSchursInequality
Date of creation	2013-03-22 15:35:25
Last modified on	2013-03-22 15:35:25
Owner	Andrea Ambrosio (7332)
Last modified by	Andrea Ambrosio (7332)
Numerical id	7
Author	Andrea Ambrosio (7332)
Entry type	Proof
Classification	msc 26D15
Classification	msc 15A42