Weyl’s inequality

Let $A$ and $E$ be two $n\times n$ Hermitian matrices, with $E$ positive semidefinite.

Let ${\lambda }_i(A)$, ${\lambda }_i(A+E)$, $1\le i\le n$ be the eigenvalues of $A$ and $A+E$ respectively, ordered in such a way that

$$|{\lambda }_1|\le |{\lambda }_2|\le \mathrm{\cdots }\le |{\lambda }_n|.$$

Then

$${\lambda }_i(A)\le {\lambda }_i(A+E).$$