The objective is as follows:


$\min\_{\mathbf{F}} a Tr(\mathbf{F} \mathbf{F}^H) - Re\{\mathbf{b}\mathbf{F}^H \mathbf{C} \mathbf{F} \mathbf{d}\}$


$s.t.\ \ \ Tr(\Sigma \mathbf{F} \mathbf{F}^H)<p$


where $a$ and $p$ are scalars, $\mathbf{b}$ is $1 \times N$ real vector, $\mathbf{d}$ is $N\times 1$ real vector, $\mathbf{C}$ is $N \times N$ Hermitian transpose matrix, $\mathbf{F}$ is $N \times N$, $\Sigma$ is $N\times N$.


It is obvious that if all the variables are scalars, the optimal $F^\ast$ is either $0$ or $p$. So how about the matrix case?


Thanks!



optimization problem, any solution?
there is a solution to your problem. 
First of all, let us consider the problem 
$$\mathcal{P}:\\ z : = \max\_{\mathbf{F}} Re\{\mathbf{b}\mathbf{F}^H \mathbf{C} \mathbf{F} \mathbf{d}\} {\rm ~~s.t.~~~} Tr(\mathbf{F} \mathbf{F}^H) \leq p. $$


Let $\mathbf{F}\_\star$ the solution to the above problem. It is clear that if $ z < a Tr(\mathbf{F}\_\star \mathbf{F}\_\star^H) $, then $\mathbf{F} = \mathbf{0}$ is the solution to your problem. Otherwise, $\mathbf{F}\_\star$ is the also the solution to your problem and $Tr(\mathbf{F}\_\star {\mathbf{F}\_\star}^H) = p$. 


Now have a deeper look at $\mathcal{P}$. As $\mathbf{C}$ is a Hermitian matrix, there exists the eigendecomposition $\mathbf{C} = \mathbf{U} \Sigma \mathbf{U}^H $, where $\mathbf{U}$ is an orthonormal matrix and $\Sigma $ is a diagonal matrix with real-valued entries $\sigma\_1,\dots,\sigma\_N$. For simplification, and without loss of generality we assume that $\mathbf{F} = \mathbf{U} \mathbf{H}$. Then, we can rewrite $\mathcal{P}$ as
$$ \mathcal{Q}: \\ 
\max\_{\mathbf{H}} Re\{\mathbf{b}\mathbf{H}^H \Sigma \mathbf{H} \mathbf{d}\} {\rm ~~s.t.~~~} Tr(\mathbf{H} \mathbf{H}^H) = p.
$$
here we have exploited the fact that $ \mathbf{U}^H \mathbf{U} = \mathbf{I}$ and $ tr( \mathbf{FF}^H)= tr( \mathbf{U}\mathbf{H} ( \mathbf{U} \mathbf{H})^H) = tr(\mathbf{HH}^H$). As $\Sigma$ is diagonal we can reformulate the cost function as
$ Re\{\mathbf{b}\mathbf{H}^H \Sigma \mathbf{H} \mathbf{d}\} = Re\{\sum\_{n=1}^N \sigma\_i \mathbf{b} \mathbf{h}\_n^H \mathbf{h}\_n \mathbf{d} \} = \sum\_{n=1}^N Re\{\sigma\_n \mathbf{h}\_n \mathbf{d} \mathbf{b} \mathbf{h}\_n^H \}$, where $\mathbf{h}\_n$ is the $n$th column of $\mathbf{H}$. Further, we see that $Re\{\sigma\_n \mathbf{h}\_n \mathbf{d} \mathbf{b} \mathbf{h}\_n^H \} = \mathbf{h}\_n(\sigma\_n\frac{1}{2}(\mathbf{d} \mathbf{b} +\mathbf{b} \mathbf{d})) \mathbf{h}\_n^H$. Defining the matrix $\mathbf{Q}\_n:= \sigma\_n\frac{1}{2}(\mathbf{d} \mathbf{b} +\mathbf{b}^H \mathbf{d}^H)$, we can finally rewrite $\mathcal{Q}$ as  

$$ \mathcal{R}: \\ 
\max\_{\{\mathbf{h}\_n\}\_{n=1}^N} \sum\_{n=1}^N \mathbf{h}\_n\mathbf{Q}\_n \mathbf{h}\_n^H {\rm ~~s.t.~~~} \sum\_{n=1}^N \| \mathbf{h}\_n\|^2 = p.
$$
From the definition it is clear that each $\mathbf{Q}\_n$ is a Hermitian matrix of rank two. All of these matrices have the (same) two eigenvectors $\mathbf{q}\_1$ and $\mathbf{q}\_2$ with (different ) eigenvalues $\lambda\_{1,n}$ and $\lambda\_{2,n}$. It is clear that vector components in $\mathbf{h}\_n$ that are orthogonal to $\mathbf{q}\_1$ and $\mathbf{q}\_2$ have no influence on the cost function of $\mathcal{R}$ but increase the constraint value. Therefore, we assume without loss of optimality that $\mathbf{h}\_n = c\_{1,n}\mathbf{q}\_1 +c\_{2,n} \mathbf{q}\_2 $. Then $\mathcal{R} $ boils down to
$$ 
\max\_{\{{c}\_{1,n},{c}\_{2,n}\}\_{n=1}^N} \sum\_{n=1}^N (\lambda\_{1,n}|c\_{1,n}|^2+\lambda\_{2,n}|c\_{2,n}|^2) {\rm ~~s.t.~~~} \sum\_{n=1}^N (|c\_{1,n}|^2+|c\_{2,n}|^2) = p.
$$
**Solution**


The solution to the above problem is simple: Let $i\_\star,n\_\star = \arg \max \lambda\_{i,n}$, then $|c\_{i\_\star,n\_\star}|^2= p $ and $|c\_{i,n}|^2= 0 $. Consequently, the solution matrix $\mathbf{H}\_\star$ has only one non-zero row (the $n\_\star$th row) which is given by $(p)^{0.5} \mathbf{q}\_{i\_\star}$. 


Consequently, $\mathbf{F}\_\star =\mathbf{U} \mathbf{H}\_\star $ and $z = p \lambda\_{i\_\star,n\_\star} $.


All the best,
Ert


