Relationship between determinant of a symmetric PSD matrix and its diagonal Assume that we have a symmetric positive semi-definite matrix $A$ with positive diagonal entries (for instance a covariance matrix). Denote with $D(A)$ the matrix whose diagonal is equal to the diagonal of $A$ and has zeros in the off-diagonal entries. For the 2-dimensional case (e.g. 2x2 matrix) it is true that:
$$ det(A) \le det(D(A))$$
My question is if this is true in the general case for higher dimensions and how it can be proven. 
 A: Ok, so I managed to answer my question (sorry for bothering anyone). The proof goes like this:
Let the eigenvalues of $A$ be $\lambda_1, \dots, \lambda_n$ and the diagonal entries of $A$ be $d_1, \dots, d_n$. 
Since $A$ is symmetric and PSD than we can write it as $A = Q \Lambda Q^T$ where $Q$ is an orthonormal matrix. The columns of $Q$ are the eigenvectors $q_i$. Let's call the rows of $Q$ with $\tilde{q}_i$. Note that since $Q$ is orthonormal it follows that $\tilde{q}_i^T\tilde{q}_i = 1$ and $\tilde{q}_i^T\tilde{q}_j=0$ for $i \not = j$. We have that:
$$d_i = \tilde{q}_i^T \Lambda \tilde{q}_i $$
For simiplicity we will analyze the $\log det(D(A))$:
$$ \log det D(A) = \sum_i \log d_i = \sum_i \log \tilde{q}_i^T \Lambda \tilde{q}_i \ge \sum_i \tilde{q}_i^T (\log \Lambda) \tilde{q}_i  $$
$$ = \sum_i Trace[ (\log \Lambda) \tilde{q}_i \tilde{q}_i^T ] = \sum_i Trace[ (\log \Lambda) Q^T Q] = Trace[\log \Lambda] $$
$$ = \log det(A) $$
The inequality at the end of the first line follows by Jensen's inequality, since $\sum_k [\tilde{q}_i]_k^2 = 1$.
