Assume that we have the following equation to solve
$$\sum\_{\ell=1}^L A\_\ell X\_{\ell} B\_{\ell} =0$$ over complex matrices 
where each $A\_{\ell}$ is a given $m\times n$ matrix, each $B\_{\ell}$ is a given $n\times t$ matrix, and we are solving for the $n\times n$ matrices $X\_\ell$. $\{A\_{\ell}\}$ satisfy
$$\sum\_{\ell=1}^L A\_\ell A\_\ell^\* = I.$$


Also, we have that $t\geq n$ and $m\geq n$, and we assume here that $L n^2 = mt+1$, so that we have exactly one more variable than we have equations to satisfy.


I would like to know conditions on $\{A\_{\ell}\}$ and $\{B\_{\ell}\}$ for the solution to be full-rank matrices $\{X\_{\ell}\}$.



Conditions for a certain matrix equation to have a full rank solution
We ignore the supplementary condition $\sum\_lA\_lA\_l^\*=I$. Adding conditions 24 hours after the first post is making fun of the world.


We assume that $n,m,t$ are fixed positive inytegers s.t.$m,t\geq n,Ln^2>mt$ and that the $(A\_l[i,j]),(B\_l[i,j])$ are $d=Ln(m+t)$ complex parameters (they are elements of a transcendental extension of $\mathbb{Q}$ and they are mutually transcendental over $\mathbb{Q}$). We consider the equation in the unknowns $(X\_l)$


$(\*)$ $\sum\_lA\_lX\_lB\_l=0$.


$\textbf{Proposition}$. The set $Z$ of $(A\_l[i,j]),(B\_l[i,j])$ s.t. $(\*)$ admits a solution $(X\_l)$ composed of invertible matrices is Zariski open dense in $\mathbb{C}^d$.


$\textbf{Proof}$. $(\*)$ is a linear equation with matrix (if we stack row by row the vectors into matrices)


$M=[A\_1\otimes B\_1^T,\cdots,A\_L\otimes B\_L^T]$.


Note that $rank(M)<mt$ can be written as algebraic relations linkink the parameters. Now, consider a solution $(X\_l)$; generically it depends on $Ln^2-mt$ free variables $(u\_j)$; assume that there is $i$ s.t. $\det(X\_i)=0$ for any choice of these free variables. $\det(X\_i)$ is a polynomial in the $(u\_j)$ that is identically $0$; that is equivalent to say that all the coefficients of this polynomial are $0$; note that these coefficients are rational fractions in the parameters.


Finally, it remains to show that $Z\not= \emptyset$. Let $A\_l=\begin{pmatrix}I\_n\\0\end{pmatrix},B\_l=\begin{pmatrix}I\_n&0\end{pmatrix}$. Then $\sum\_l A\_lB\_lX\_l=\begin{pmatrix}\sum\_lX\_l&0\\0&0\end{pmatrix}=0$ iff $\sum\_lX\_l=0$. Since $L\geq 2$, one can always choose the $(X\_l)$ as invertible matrices. $\square$


$\textbf{Remark}.$ That implies that if we randomly choose the $(A\_l),(B\_l)$ , then, with probability $1$, there exists an "invertible" solution of $(\*)$. 


PS. Your relation $(\*\*)$ $\sum\_lA\_lA\_l^\*=I$ is not algebraic over $\mathbb{C}$ but only over $\mathbb{R}$. Thus, it is more difficult to prove a variant of the previous proposition including the condition $(\*\*)$. In particular, it does not suffice to prove that the "new" $Z$ is non-void.


