# 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

$=\operatorname{Tr}(D^{H}D)+\sum_{i

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

Title proof of Schur’s inequality ProofOfSchursInequality 2013-03-22 15:35:25 2013-03-22 15:35:25 Andrea Ambrosio (7332) Andrea Ambrosio (7332) 7 Andrea Ambrosio (7332) Proof msc 26D15 msc 15A42