proof of Weyl’s inequality

Let ${\lambda }_i$ be the i-th eigenvalue of $A+E$. Then, by the Courant-Fisher min-max theorem and being $x^{H}Ex\ge 0$ by hypothesis, we have:

${\lambda }_i(A+E)=\underset{S,dimS=i}{\max }\underset{\parallel x\parallel \ne 0}{\min }\frac{x^H(A+E)x}{x^{H}x}=$
$=\underset{S,dimS=i}{\max }\underset{\parallel x\parallel \ne 0}{\min }\left(\frac{x^{H}Ax}{x^{H}x}+\frac{x^{H}Ex}{x^{H}x}\right)\ge \underset{S,dimS=i}{\max }\underset{\parallel x\parallel \ne 0}{\min }\frac{x^{H}Ax}{x^{H}x}={\lambda }_i(A)$.

Title	proof of Weyl’s inequality
Canonical name	ProofOfWeylsInequality
Date of creation	2013-03-22 15:33:39
Last modified on	2013-03-22 15:33:39
Owner	Andrea Ambrosio (7332)
Last modified by	Andrea Ambrosio (7332)
Numerical id	9
Author	Andrea Ambrosio (7332)
Entry type	Proof
Classification	msc 15A42