# proof of Schur’s inequality

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

$$

$$

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

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 |