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.



if Y-X is positive semi-definite, are the eigenvalues of Y bigger? 
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.


