Linear algebra problem involving concatenation of matrices I have a matrix problem I don't know how to solve because I can't seem to express it as a multiplication of matrices.
Suppose I have the following 4x4 matrix $A$:
$$\begin{bmatrix}0&0&1&0\cr 0&0&1&1\cr 1&0&0&1\cr 0&1&0&0\end{bmatrix}$$
The problem is to find a 2x4 matrix $B$ such that the 4x4 matrix $C = \begin{bmatrix}B \cr BA\end{bmatrix}$ has full rank.
For example:


*

*$B_1 = \begin{bmatrix}0&0&0&1\cr 0&0&1&0\end{bmatrix}$ is a solution, because $\begin{bmatrix}B \cr BA\end{bmatrix} = \begin{bmatrix}0&0&0&1\cr 0&0&1&0\cr 0&1&0&0\cr 1&0&0&1\end{bmatrix}$ has full rank

*$B_2 = \begin{bmatrix}0&0&0&1\cr 0&1&0&0\end{bmatrix}$ is not a solution, because $\begin{bmatrix}B \cr BA\end{bmatrix} = \begin{bmatrix}0&0&0&1\cr 0&1&0&0\cr 0&1&0&0\cr 0&0&1&1\end{bmatrix}$ has rank 3
Is there a way to express this problem more naturally, so it has a solution using conventional techniques of linear algebra, rather than trial and error?

p.s. note: this is only an example, which is easy to do by hand; I'm wondering if there are any general methods for problems of this type, where the matrices in question might be much larger. I was hoping to be able to transform this equation into some kind of decomposition or nullspace problem.
For example, if it helped to calculate $D = \begin{bmatrix}B & BA\end{bmatrix}$ then this can be written as $D = BA'$ where $A' = \begin{bmatrix}I & A\end{bmatrix}$ and maybe I could use the structure of $A'$ to determine $B$ more directly.
(I'm interested in solutions with scalar calculation in $GF(2)$ but the math works out the same for these examples whether it's in $GF(2)$ or in "regular" algebra)
 A: Let $B$ be composed of the row vectors $(b_1,b_2,b_3,b_4)$ and $(c_1,c_2,c_3,c_4)$. 
Then the matrix $\begin{pmatrix}B\\BA\end{pmatrix}=\begin{pmatrix}b_1 & b_2 & b_3 & b_4 \\ c_1 & c_2 & c_3 & c_4 \\ b_3 & b_4 & b_1+b_2 & b_2+b_3 \\ c_3 & c_4 & c_1+c_2 & c_2+c_3\end{pmatrix}$ has full rank if and only if its determinant (which gives you a constraint for the variables) is nonzero.
A: This is related to the observability matrix, which if it is full rank also satisfies the Hautus lemma. This states that the observability matrix would be full rank iff
$$
\text{rank}\begin{bmatrix}A^\top - \lambda\,I & B^\top\end{bmatrix} = n\quad \forall\ \lambda\in\mathbb{C}
$$
where $A$ is a $n$ by $n$ matrix. The first term only loses rank when $\lambda$ is an eigenvalue of $A$. In those cases the whole matrix still has full rank if the null space of $B^\top$ does not coincide with the null space of $A^\top - \lambda\,I$. In order words in your case the span of the of $B^\top$ combined with any $n/2$ eigenvectors of $A^\top$ should always contain $n/2$ linearly independent vectors (assuming that the eigenvalues of $A$ do not have a geometric multiplicity greater then one).
For example setting each column of $B^\top$ equal to the sum of two of the eigenvectors, such that each eigenvectors is only used for one of the columns of $B^\top$.
