Schur’s inequality
Theorem (Schur’s inequality) Let $A$ be a square $n\times n$ matrix with real (or possibly complex entries). If ${\lambda}_{1},\mathrm{\dots},{\lambda}_{n}$ are the eigenvalues^{} of $A$, and $D$ is the diagonal matrix^{} $D=\mathrm{diag}({\lambda}_{1},\mathrm{\dots},{\lambda}_{n})$, then
${\parallel D\parallel}_{F}$ | $\le $ | ${\parallel A\parallel}_{F},$ |
where $\parallel \cdot {\parallel}_{F}$ is the Frobenius matrix norm. Equality holds if and only if $A$ is a normal matrix^{}.
References
- 1 V.V. Prasolov, Problems and Theorems in Linear Algebra, American Mathematical Society, 1994.
Title | Schur’s inequality |
---|---|
Canonical name | SchursInequality |
Date of creation | 2013-03-22 13:43:30 |
Last modified on | 2013-03-22 13:43:30 |
Owner | matte (1858) |
Last modified by | matte (1858) |
Numerical id | 14 |
Author | matte (1858) |
Entry type | Theorem |
Classification | msc 26D15 |
Classification | msc 15A42 |
Related topic | TraceOfAMatrix |
Related topic | WielandtHoffmanTheorem |
Related topic | FrobeniusMatrixNorm |