# 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 $$\{X_{\ell}\}$$.

• Nice question, I am curious too. By the way, do you want every solution to be a full-rank matrix? This is impossible since the matrix whose entires are all zero satisifies the equation trivially. I think you wanted to ask about conditions in order a full rank matrix solution to exist, right? Commented Dec 13, 2019 at 10:04
• Correct, it is enough that there exist one solution that full rank matrices. Commented Dec 13, 2019 at 10:58

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 $$rank(M) can be written as algebraic relations linkink the parameters. Now, consider a solution $$(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.

• Thanks for your post - and sorry for changing the question so late. When writing the question I removed the condition since I felt it was not so restrictive, but later realised it was. Commented Dec 14, 2019 at 19:54
• You are welcome. Commented Dec 14, 2019 at 20:39