TITLE: Why does ${\lambda _i}(A) \ge {\lambda _i}(B)$?
QUESTION [5 upvotes]: Let $A,B \in {M_n}$ are Hermitian and $A-B$ has only nonnegative eigenvalues.Why does ${\lambda _i}(A) \ge {\lambda _i}(B)$  (for $i=1,2,\ldots,n$) ?

REPLY [3 votes]: This is a consequence of the Monotonicity Theorem, which says that if $A$ and $B$ are Hermitian and $B$ is positive semi-definite, then
$$\lambda_i(A+B) \ge \lambda_i(A).$$
Indeed, the Montonicity Theorem itself is a consequence of Weyl's inequalities, which says if $A,B$ are $n\times n$ Hermitian, then
$$\lambda_i(A) + \lambda_1(B) \ge \lambda_i(A+B) \ge \lambda_i(A) + \lambda_n(B).$$
We can see that $B$ being positive semi-definite means that $\lambda_n(B) \ge 0$ so that the monotonicity theorem follows immediately from Weyl's inequalities.
For your result, we simply apply the monotonicity theorem to $B$ and $A-B$ (which is positive semi-definite) to get
$$\lambda_i(A) = \lambda_i(B + (A-B)) \ge \lambda_i(B).$$