How to solve the optimization problem $\mathbf{F}=\arg\min_{\mathbf{F}} tr((\mathbf{G}^H\mathbf{F}^H\mathbf{F}\mathbf{G}+\alpha\mathbf{I})^{-1})$

Question

I have an optimization problem, which tries to minimize the trace $\mathrm{trace}((\mathbf{G}^H\mathbf{F}^H\mathbf{F}\mathbf{G}+\alpha\mathbf{I})^{-1})$ with respect to $\mathbf{F}$, where $\mathbf{G}$ is a given complex matrix of size $N\times K$ ($N>K$), and $\mathbf{F}$ is of size $L\times N$ ($L<N$) which has the power constraint $||\mathbf{F}||^2_{\text{F}}\leq K$ (frobenius norm), $\alpha$ is a positive constant, $\mathbf{I}$ is an identity matrix of size $K\times K$. That is, $$ \begin{array} \,\min_{ F\in\mathbb{C}^{L\times N} } & \mathrm{trace}((\mathbf{G}^H\mathbf{F}^H\mathbf{F}\mathbf{G}+\alpha\mathbf{I})^{-1}) \\ \mathrm{such \;\;that} %& %G\in \mathbb{C}^{N\times K} %\\ & ||\mathbf{F}||^2_{\text{F}}\leq K %& \end{array} $$

Are there any experts aware of this optimization problem? Thanks in advance!

Answer 1

Remark 1. Note that if $F$ realizes the minimum of the studied function, then, for every complex number $u$ of modulus $1$, $uF$ realizes also the minimum.

Remark 2. $F=0$ realizes the maximum $\dfrac{n}{\alpha}$ of the studied function.

Lemma: Let $Z\in M_{N,L}(\mathbb{C})$ s.t., for every $H\in M_{L,N}(\mathbb{C})$, $Re(tr(HZ))=0$. Then $Z=0$.

Proof. Let $H=H_1+iH_2,Z=Z_1+iZ_2$ where $H_i,Z_i$ are real matrices. For every $H_1,H_2$, $tr(H_1Z_1-H_2Z_2)=0$. Putting $H_2=0,H_1=Z_1^*=Z_1^T$ and, in a second time, $H_1=0,H_2=Z_2^*=Z_2^T$, we deduce $Z_1=Z_2=0$ and, consequently, $Z=0$.

The matrix $U=G^*F^*FG+\alpha I_k$ is a $K\times K$ hermitian $>0$ matrix and $U^{-1}$ too. We study the function $f:F\in M_{L,N}(\mathbb{C})\rightarrow tr(U^{-1})\in (0,+\infty)$ under the condition $(*)$: $\phi(F)=tr(FF^*)-\tau\leq 0$. We consider that $F\in \mathbb{R}^{2LN}$ and we calculate the derivative of $f$ with respect to real variables; in particular, the derivative of $z=x+iy\in\mathbb{C}\sim\mathbb{R}^2\rightarrow \bar{z}$ is $h\in\mathbb{C}\rightarrow \bar{h}$ (modulo the identification between $\mathbb{C}$ and $\mathbb{R}^2$). Note that the conjugation is not differentiable with respect to a complex variable.

The required derivative is $Df_F:H\in M_{L,N}(\mathbb{C})\rightarrow -tr(U^{-1}(G^*H^*FG+G^*F^*HG)U^{-1})=-2Re(tr(U^{-1}G^*F^*HGU^{-1})$ or $Df_F(H)=-2Re(tr(GU^{-2}G^*F^*H))$.

Case 1. The condition $(*)$ is open, that is $\phi(F)<0$. For every $H$, $Df_F(H)=0$; according to Lemma, that implies $GU^{-2}G^*F^*=0,tr(FF^*)<\tau$.

Case 2. The condition $(*)$ is closed, that is $\phi(F)=0$. The Lagrange condition is: there is $\lambda\in\mathbb{R}$ s.t. $Df_F-\lambda D\phi_F=0$. Here $D\phi_F(H)=tr(HF^*+FH^*)=2Re(tr(F^*H))$. According to Lemma, this condition can be written $GU^{-2}G^*F^*-\lambda F^*=0,tr(F^*F)=\tau$.

Beware, $U$ depends on $F$; consequently, there is no closed form of the solution in Case 1 and Case 2.