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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 eigen-values of the two matrices?

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closed as off-topic by Will Jagy, Carlo Beenakker, David White, Suvrit, Vidit Nanda Oct 3 '13 at 0:15

This question appears to be off-topic. 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
If this question can be reworded to fit the rules in the help center, please edit the question.

What does hold is that $\|A^*\| \le \|A\|$ for every unitarily invariant norm $\|\cdot\|$; in fact the eigenvalues of $A^*$ are majorized by those of $A$. – Suvrit Oct 2 '13 at 21:54
would you give some proof in a separate answer? – behrad mahboobi Oct 2 '13 at 22:17
Currently I see your question heading towards closure. You might want to provide some background to your question, and also as my comment shows, the desired inequality cannot hold. However, some interesting norm inequalities between these two nice matrices can be shown---along with obvious stuff like they have the same trace etc. Once you clarify your question a bit, and no one else answers, then I'll spend some time to expand my comment... – Suvrit Oct 2 '13 at 22:45
Take $A$ to be the all $1$s matrix. – Suvrit Oct 2 '13 at 23:38
Correct!!! Thank you – behrad mahboobi Oct 2 '13 at 23:41
up vote 1 down vote accepted

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.

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thank you for your post. I have accepted your answer however you are welcome to write more about this post! – behrad mahboobi Oct 3 '13 at 0:09

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