Expected value of a random matrix and the expected value of its eigenvalues For $M$ being a $p \times p$ symmetric positive definite matrix. If $E(M) = A$, does that imply $E(\lambda_i) = l_i$, for $i = 1,...,p$, where $\lambda_i$ and $l_i$ are the $i$th eigenvalues of $M$ and $A$, respectively? Thanks. 
 A: (edited) I don't think it holds. Take $2\times 2$ diagonal matrices with the two diagonal components $X$ and $Y$ independent and uniformly distributed on $[0,1]$. Then $E(M)=\mathrm{diag}(1/2, 1/2)$, but $E(\lambda_1)=E\, \min\{X,Y\}$ is less than $1/2$ and $E(\lambda_2)=E \,\max\{X,Y\}$ more than $1/2$.
A: The question looks ambiguous to me because, the eingenvalues of a matrix don't have an intrisic order, so you cannot directly compare the $i$th eigenvalue of $A$ with that of $M$. 
Unless you are assuming that the eigenvalues are to be ordered by magnitude, in which case the property is obviously false (see Peter answer, for example).
We could ask if we can define some eingenvalues order so that the property holds. Again, that's false. It would be true if you could write each eigenvalue as a linear function of the matrix entries (not true), or if the space of random matrices share a common (constant) eigenvector matrix $P$. In the later case, one could write $\Lambda_M = P^t M P$ so that $E(\Lambda_M)=P^t E(M) P=P^t A P= \Lambda_A$. But, again, this is not true in general.
A weaker -but true- property is that the expected value of the average value of all the eigenvalues of $M$ is equal to the average value of the eigenvalues of $A$. This can be deduced from the property $\sum \lambda_i = tr(M) $ which is indeed a linear function of the matrix entries. Alternatively: the expected value of a random (chosen at random) eigenvalue of the random matrix $M$ is equal to the expected value of a random eigenvalue taken from $A=E(M)$.
