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


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

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

But I have no idea how to solve this, could someone please help me with that?

  • $\begingroup$ You have $\binom{N+1}{2}$ equations in $N M$ unknowns. Please define "solve". $\endgroup$ Apr 20, 2022 at 6:56
  • $\begingroup$ 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. $\endgroup$
    – richrow
    Apr 20, 2022 at 7:06
  • $\begingroup$ 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 $\overline{B}$. $\endgroup$
    – user1551
    Apr 20, 2022 at 7:09
  • $\begingroup$ @RodrigodeAzevedo Yes, "conjugate symmetry" means Hermitian. "solve" means to find the closed form expression of $\mathbf{X}$. $\endgroup$
    – ZYX
    Apr 20, 2022 at 7:09
  • $\begingroup$ @user1551 Thanks, I have edited it again. $\endgroup$
    – ZYX
    Apr 20, 2022 at 7:13

1 Answer 1


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.


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