Random matrices: stochastic dominance of trace and determinant For this question, if $X$ and $Y$ are 2 real random scalars, then $X$ has first-order stochastic dominance over $Y$ if for any $x$, $\Pr[X\leq x]\leq\Pr[Y\leq x]$. Let's write $X\succeq Y$.

Suppose $A$ and $B$ are $n\times n$ random positive semidefinite real matrices and I'm interested in sufficient conditions that allow me to say $\operatorname{tr}(A)\succeq\operatorname{tr}(B)$ or $\det(A)\succeq\det(B)$. In particular, can I say something if I know $c'Ac\succeq c'Bc$ for all $n\times 1$ nonstochastic $c$?

 A: For the trace, you can use $c$ equal to unit vectors to pick out the diagonal elements of the matrices.  Each diagonal element of (A) has stochastic dominance over the corresponding element of (B), the sum of r.v.s from (A) will have stochastic dominance over the sum of r.v.s from (B).
For the determinant, you can use the fact that $\det\left(M\right)=e^{\mathrm{Tr}\left(\log\left(M\right)\right)}$ (you might have to work a bit harder to deal with the case where you have zero eigenvalues).  If we take $c$ to be the various elements of an eigenvector (or Jordan) basis for $B$, then each eigenvalue of $B$ is stochastically dominated by a diagonal element of $A$ in this basis.  These elements of $A$ sum to $\mathrm{Tr}\left(A\right)$.  Since $\log$ is concave, the sum of the $\log$s of these elements should be stochastically dominated by the sum of the $\log$s of the eigenvalues of $A$.  (Please verify the details on this—I haven't checked everything very carefully and I'm leaving the details of the proof to you.)  So $\mathrm{Tr}\left(\log\left(A\right)\right)\succeq\mathrm{Tr}\left(\log\left(B\right)\right)$.  Since $e^{x}$ is an increasing function, $e^{\mathrm{Tr}\left(\log\left(A\right)\right)}\succeq e^{\mathrm{Tr}\left(\log\left(B\right)\right)}$, so $\det A\succeq\det B$.
