# How to solve the matrix equation $\mbox{tr}(\mathbf{X}\mathbf{X}^H)\overline{\mathbf{B}}=\mbox{tr}(\mathbf{X}\mathbf{X}^H\mathbf{B})\mathbf{I}$?

I want to solve the following equation for $$\mathbf{X}\in\mathbb{C}^{N\times M}$$, with $$M < N$$:

$$\mbox{tr}(\mathbf{X}\mathbf{X}^H)\overline{\mathbf{B}}=\mbox{tr}(\mathbf{X}\mathbf{X}^H\mathbf{B})\mathbf{I}$$

where $$\mathbf{B}$$ is a known $$N\times N$$ Hermitian positive semidefinite matrix, and $$\mathbf{I}$$ is the identity matrix. $$\overline{\mathbf{B}}$$ means the complex conjugate of $$\mathbf{B}$$.

I have tried to express $$\mathbf{B}$$ as $$\mathbf{B} = \mathbf{b}\mathbf{b}^H$$, and got the following equation

$$\mbox{tr}(\mathbf{X}\mathbf{X}^H)\overline{\mathbf{b}\mathbf{b}^H}=\mbox{tr}(\mathbf{X}\mathbf{X}^H\mathbf{b}\mathbf{b}^H)\mathbf{I} = \mathbf{b}^H\mathbf{X}\mathbf{X}^H\mathbf{b}\mathbf{I}$$

but still hard to solve.

I also tried to vectorize the matrix, and got

$$\mbox{vec}^H(\mathbf{X})\mbox{vec}(\mathbf{X})\overline{\mathbf{b}\mathbf{b}^H}=\mbox{vec}^H(\mathbf{X})(\mathbf{X}^H \otimes \mathbf{I})\mbox{vec}(\mathbf{b}\mathbf{b}^H)\mathbf{I} = \mbox{vec}^H(\mathbf{X})\mbox{vec}(\mathbf{b}\mathbf{b}^H \mathbf{X}^H)\mathbf{I}$$

• You have $\binom{N+1}{2}$ equations in $N M$ unknowns. Please define "solve". Apr 20, 2022 at 6:56
• It should be clear that either $\mathbf{B}=\lambda\cdot \mathbf{I}$ for $\lambda\in\mathbb{R}$ (and $\mathbf{X}$ is arbitrary) or $\mathbf{X}=0$). Note that your equation is of the form $a\cdot \mathbf{B}^*=b\cdot \mathbf{I}$, where $a$ and $b$ are some constants. Apr 20, 2022 at 7:06
• Your symbols are confusing. In linear algebra literature, both $B^\ast$ and $B^H$ mean the conjugate transpose of $B$. What is the $B^\ast$ in your question? If you mean complex conjugate, please change it to the standard notation $\overline{B}$. Apr 20, 2022 at 7:09
• @RodrigodeAzevedo Yes, "conjugate symmetry" means Hermitian. "solve" means to find the closed form expression of $\mathbf{X}$.
– ZYX
Apr 20, 2022 at 7:09
• @user1551 Thanks, I have edited it again.
– ZYX
Apr 20, 2022 at 7:13

Clearly $$X=0$$ is a solution.
Now suppose $$X\ne0$$. Then $$XX^H$$ is a nonzero positive semidefinite matrix. Therefore $$\operatorname{tr}(XX^H)>0$$ and the equation implies that $$\overline{B}=kI$$ where $$k=\frac{\operatorname{tr}(XX^HB)}{\operatorname{tr}(XX^H)}$$. The equation can therefore be rewritten as $$\operatorname{tr}(XX^H)\overline{k}I=\operatorname{tr}(kXX^H)I$$ and further be rewritten as $$\operatorname{tr}(XX^H)\overline{k}I=\operatorname{tr}(XX^H)kI$$. Since $$\operatorname{tr}(XX^H)>0$$, it is solvable only if $$k$$ is real. If this is the case, every nonzero $$X$$ is a solution.
So, in summary, every (zero or nonzero) $$X$$ is a solution when $$B$$ is a real scalar multiple of $$I$$, or $$X=0$$ is the only solution otherwise.