Solve a trace equality

Question

I have the following trace equality:

$\operatorname{tr}(XA)+BX^TC=D $

Where $X\in\mathbb{R}^{m\times r}$, $A\in\mathbb{R}^{r\times m}$, $B\in\mathbb{R}^{1\times r}$, $C\in\mathbb{R}^{m\times 1}$, $D\in\mathbb{R}$.

Is there a way to find $X$?

Answer 1

Note that $\operatorname{tr}(XA)+BX^TC=\operatorname{tr}\left((A^T+CB)^TX\right)$.

If $A^T+CB=0$, the equation is solvable only when $D=0$, and in this case every matrix $X\in\mathbb R^{m\times r}$ is a solution.

If $A^T+CB\ne0$, the general solution to $\operatorname{tr}\left((A^T+CB)^TX\right)=D$ is given by $$ X=\frac{D(A^T+CB)}{\operatorname{tr}\left((A^T+CB)^T(A^T+CB)\right)}+Y, $$ where $Y$ is any matrix such that $\operatorname{tr}\left((A^T+CB)^TY\right)=0$. So, one particular solution is obtained by putting $Y=0$ in the above.

Answer 2

No way. You have a single linear equation for a matrix with $mr$ entries. That leaves an (at least) $(mr-1)$-dimensional space of solutions.