Conditions for a certain matrix equation to have a full rank solution 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 $\{X_{\ell}\}$.
 A: 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.
