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I am facing this question. I need to optimize the precoding matrix to maxmize the channel capacity, which can be derived as follows $$\max_{\textbf{B}}\ \text{tr}\left(\textbf{A}\textbf{B}\right)$$ $$\text{s.t.}\ \ \ \ \text{tr}(\textbf{B})\leq p$$ $\textbf{A}=\textbf{HH}^H$, where $\textbf{H}\in\mathbb{C}^{a\times a}$ is the channel state information, and $\textbf{B}=\textbf{FF}^H$, where $\textbf{F}\in\mathbb{C}^{a\times a}$ is the precoding matrix. The max tranmit power is $p$. $a$ is the number of receive atennas.

I have inspired by the comment under the question and know the solution is $\textbf{B}=p\textbf{v}_1\textbf{v}_1^H$, where $\textbf{v}_1$ is the first right singular vector of $\textbf{A}$. So, the optimal precoding matrix $\textbf{F}=\textbf{v}_1\textbf{x}$, where $\textbf{x}$ satisfies $||\textbf{x}||^2_2=p$. But how can I prove it? Why is this the upper bound?

I have corrected this question many times and I don't know why I can't reopen it. I think I've done what was asked of me. Though I have inspired by the comment and know the answer of the question I raised, but I still being asked to revise. If the admins don't allow to reopen it, can I delete it? I don't know how to do.

RACHpreamble
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    Using matrix square roots of $A$ and $B$ I believe the problem can be converted to something like this – whpowell96 Dec 02 '24 at 03:49
  • I am not sure what positive definite means in the complex field. I think you are making this much more complicated than it needs to be. Write the problem as $\sup_{\langle I, B \rangle \le \rho} \langle A, B \rangle$. By analogy with $\mathbb{C}^n$ you should be able to conjure up a solution. Note that for 'most' $A$, the program is unbounded. – copper.hat Dec 02 '24 at 04:55

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If I'm not missing anything,

The algorithm is given an input of $A$, a positive semi-definite symmetric matrix, as well as a real $p>0$.

It has to find the maximum value of $\text{tr}(AB)$ as $B$ ranges over the positive semi-definite symmetric matrices with a trace of at most $p$.

But $\text{tr}(AB)=\text{tr}(U^*AUU^*BU)$ and $\text{tr}(B)=\text{tr}(U^*BU)$, where $U$ is an orthonormal matrix such that $U^*AU$ is diagonal.

So you look at the diagonal entries of the diagonalisation of $A$, we can denote them $a_1$ through $a_k$. By the premise of the problem, these terms are non-negative.

So, the only task is to maximise $\sum a_i\beta_i$, where $\beta_i$ refers to the diagonal entries of $U^*BU$. The only constraint is that these terms are non-negative and sum to $p$.

Use Cauchy-Schwarz to tell you how to make the choice of $\beta_i$.

Chris Sanders
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