if Y-X is positive semi-definite, are the eigenvalues of Y bigger?  So $X$ and $Y$ are Hermitian matrices (or just symmetric real) of size $n$ by $n$ and suppose $Y\succeq X$, namely $Y-X$ is positive-semidefinite. Now write the eigenvalues of $Y$ as $\alpha_1\leq\ldots\leq \alpha_n$, and the eigenvalues of $X$ as $\beta_1\leq\ldots\leq \beta_n$. Is is necessarily true that $\alpha_i\geq\beta_i$ for all $i$?
I might be able to solve this myself (although with time I am less sure), but it should be be much easier for whoever already knows the answer. A quick reference would do, thanks.
 A: This is true and well known. By the minimax principle, $\alpha_k$ is the minimum over all $k$-dimensional subspaces of the norm of the quadratic form $v\mapsto(v,Yv)$ restricted to the subspace. And similarly for $\beta_k$ and $(v,Xv)$. Since $(v,Yv)\ge(v,Xv)$ for every vector $v$, the same inequality holds for the norms of restrictions to subspaces and hence for eigenvalues.
I think any textbook covering the minimax principle has this corollary.
A: At this point I find it worth mentioning the following facts. Let $X$ and $Y$ be arbitrary square complex matrices. Let $|X|=(X^*X)^{1/2}$ be the matrix absolute value. Then, we may have the following inequalities (in decreasing order of strength, i.e., a higher one in the list implies a lower one):


*

*$|X| \succeq |Y|$ 

*$\sigma_i(X) \ge \sigma_i(Y)$, where $\sigma_i$ is the $i$th singular value

*$\prod_{j=1}^k \sigma_j^\downarrow(X) \ge \prod_{j=1}^k \sigma_j^\downarrow(Y)$ for $k=1,2,\ldots,n$;

*$\sum_{j=1}^k \sigma_j^\downarrow(X) \ge \sum_{j=1}^k \sigma_j^\downarrow(Y)$ for $k=1,2,\ldots,n$;

*$\| X\| \ge \|Y\|$, where $\|\cdot\|$ is any unitarily invariant norm.


Actually, in the list above, $4\Leftrightarrow 5$. The original question is a special case of $1 \implies 2$, and the textbook proof is as S. Ivanov has nicely explained above.
A: No, but
$$
 \sum_{j=1}^{k} \alpha_j \geq \sum_{j=1}^{k} \beta_j
$$ 
for all $k$ as EASILY follows from the mini max principle.
