Coupled Sylvester equations Let $n \in \mathbb{N}$. Let $A,B,C$ real matrices of size $n \times n$. Let $\alpha,\beta,\gamma,\delta \in \mathbb{R}^{4}$ such that $\alpha,\beta$ are non-zero.   
I an looking for two matrices $T_1$ and $T_2$ belonging to $\mathbb{R}^{n \times n}$ such that:
\begin{eqnarray}
\alpha A T_1 + T_1 B &=& \gamma C T_2 \\ 
\beta A T_2 + T_2 B &=& \delta C T_1
\end{eqnarray}


*

*Is there any known result on when these coupled Sylvester equations admit a solution?

*If there is a solution, when is it unique (as a function of the scalars $\alpha,\beta,\gamma,\delta$) ?


Edit: as was mentioned by Carlo Beenakker, the equations admit the trivial solution $T_1 = T_2 = 0$. What if now, we assume both $T_1$ and $T_2$ non-singular?
 A: There is a simpler criterion that does not involve constructing a $2n^2 \times 2n^2$ matrix and checking its singularity (which is a costly operation).
Your equation is equivalent to the standard Sylvester equation (with rectangular unknown)
$$
M \begin{bmatrix}T_1\\T_2\end{bmatrix} + \begin{bmatrix}T_1\\T_2\end{bmatrix}  B = 0,
$$
where
$$M=
\begin{bmatrix}
\alpha A & -\gamma C\\
-\delta C & \beta A
\end{bmatrix},
$$
so you can rely on all the existing theory for the Sylvester equation. For instance, a classical result is: there is a nonzero solution to the equation iff $M$ and $-B$ have no common eigenvalues.
A: Your coupled equations have the the trivial solution $T_1=T_2=0$; coupled Sylvester equations have a different form 
$$AT_1+T_2B=C,\;\;DT_1+T_2E=F,$$
with a constant right-hand-side. For given $A,D\in\mathbb{R}^{m\times m}$, $B,E\in \mathbb{R}^{n\times n}$, $C,F\in\mathbb{R}^{m\times n}$ these have a unique solution $T_1,T_2\in\mathbb{R}^{m\times n}$ if and only if the following matrix $S$ is nonsingular:
$$S=\begin{pmatrix}
I_n\otimes A&B^T\otimes I_m\\
I_n\otimes D&E^T\otimes I_m
\end{pmatrix},$$
see for example Iterative least-squares solutions of coupled Sylvester matrix equations (2005).
A: Your equations are a linear system $L \pmatrix{T_1\cr T_2\cr} = \pmatrix{0\cr 0\cr}$, where $L$ is a linear map from $(\mathbb R^{n \times n})^2$ to itself, thus essentially a $2n^2 \times 2n^2$ matrix.  Compute its null space.  If
there are $T_1, T_2$ such that $\det(T_1)$ and $\det(T_2)$ are nonzero, this must be true for a generic member of the null space.
A: We have two coupled linear matrix equations in $\mathrm X_1, \mathrm X_2 \in \mathbb R^{n \times n}$
$$\begin{array}{rl} \alpha \, \mathrm A \mathrm X_1 + \mathrm X_1 \mathrm B &= \gamma \, \mathrm C \mathrm X_2\\ \beta \, \mathrm A \mathrm X_2 + \mathrm X_2 \mathrm B &= \delta \, \mathrm C \mathrm X_1\end{array}$$
Vectorizing both sides of both matrix equations, we obtain the following homogeneous linear system
$$\begin{bmatrix} \left((\mathrm I_n \otimes \alpha \, \mathrm A) + (\mathrm B^\top \otimes \mathrm I_n)\right) & - \mathrm I_n \otimes \gamma \, \mathrm C\\ - \mathrm I_n \otimes \delta \, \mathrm C & \left((\mathrm I_n \otimes \beta \, \mathrm A) + (\mathrm B^\top \otimes \mathrm I_n)\right) \end{bmatrix} \begin{bmatrix} \mbox{vec} (\mathrm X_1)\\ \mbox{vec} (\mathrm X_2)\end{bmatrix} = \begin{bmatrix} 0_{n^2}\\ 0_{n^2}\end{bmatrix}$$ 
One solution is $\mathrm X_1 = \mathrm X_2 = \mathrm O_n$. The rank of the $2n^2 \times 2n^2$ block matrix above tells us whether this solution is unique or not.
