Can we bound $\mathbf{A}$ with $\mathbf{A^*}$ as ${\bf{A}} \preceq {{\bf{A}}^*}$ where \begin{equation} {\bf{A}} = \left[ {\begin{array}{*{20}{c}} {{{\bf{A}}_{11}}}&{...}&{{{\bf{A}}_{1N}}}\\ \vdots & \ddots & \vdots \\ {{\bf{A}}_{1N}^H}&{...}&{{{\bf{A}}_{NN}}} \end{array}} \right],\quad {{\bf{A}}^*} = \left[ {\begin{array}{*{20}{c}} {{{\bf{A}}_{11}}}&{...}&{\bf{0}}\\ \vdots & \ddots & \vdots \\ {\bf{0}}&{...}&{{{\bf{A}}_{NN}}} \end{array}} \right] \end{equation} actually all $\mathbf{A}_{i,i}$ and $\mathbf{A}$ are positive semidefinite matrices. If the above inequality cannot hold, please remark if with some extra assumptions on $\mathbf{A}_{i,i}$, it can be made to hold? what other facts can be stated about the eigenvalues of the two matrices?

closed as offtopic by Will Jagy, Carlo Beenakker, David White, Suvrit, Vidit Nanda Oct 3 '13 at 0:15
This question appears to be offtopic. The users who voted to close gave this specific reason:
 "This question does not appear to be about research level mathematics within the scope defined in the help center." – Carlo Beenakker, David White, Suvrit, Vidit Nanda
Actually, there is a great question hiding in there that the OP could have asked. Since he does not care about norm inequalities or other majorization results, here's the trivial most counterexample. \begin{equation*} A = \begin{pmatrix} 1 & 1\\ 1 & 1 \end{pmatrix}. \end{equation*} It follows from results on matrix pinchings that \begin{equation*} \lambda(A^*) \prec \lambda(A) \end{equation*} where $\lambda(\cdot)$ is the eigenvalue map. Other more interesting things can also be said about this question, but I'll write them out only if there is more interest. EDIT. Just adding some material for the interested. Positive definite matrices with Hermitian blocks (not necessarily psd as in the OP), arise in quantum information theory and related areas. Indeed, Bourin, Lee, and Lin show in [1] that if $A$ is a positive semidefinite matrix partitioned into $N\times N$ blocks, then for all unitarily invariant norms \begin{equation*} \A\ \le \left\\sum_{i=1}^N A_{ii}\right\. \end{equation*} Their proof is based on generators of a Clifford algebra. Of course, this subject has connections to the operation of taking partial traces. [1] J.C. Bourin, E.Y. Lee, and M. Lin. Positive definite matrices with Hermitian blocks and their partial traces, arXiv, 2012. 

