# 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 $x^{H}Ex\geq 0$ by hypothesis, we have:

$\lambda_{i}(A+E)=\max\limits_{S,dimS=i}\min\limits_{\|x\|\neq 0}\frac{x^{H}(A+% E)x}{x^{H}x}=$
$=\max\limits_{S,\dim S=i}\min\limits_{\|x\|\neq 0}\left(\frac{x^{H}Ax}{x^{H}x}% +\frac{x^{H}Ex}{x^{H}x}\right)\geq\max\limits_{S,\dim S=i}\min\limits_{\|x\|% \neq 0}\frac{x^{H}Ax}{x^{H}x}=\lambda_{i}(A)$.

Title proof of Weyl’s inequality ProofOfWeylsInequality 2013-03-22 15:33:39 2013-03-22 15:33:39 Andrea Ambrosio (7332) Andrea Ambrosio (7332) 9 Andrea Ambrosio (7332) Proof msc 15A42