rank of this matrix 
*

*${\rm rank\,}\begin{bmatrix}B&AB&\cdots&A^{n-1}B\end{bmatrix}=n$ if and only if for all $\lambda\in\mathbb{C}$, rank$\,\begin{bmatrix}\lambda I_n-A&B\end{bmatrix}=n.$

*$\otimes$ is the Kronecker product

 A: A possibly useful note:
I'll denote
$$
{\rm rank\,}\begin{bmatrix}I_n\otimes A- A^{\rm T}\otimes  I_n\\B^{\rm T}\otimes I_n\end{bmatrix}=n^2 \tag{1}
$$
as Equation (1).  Note that 
$$
[I \otimes A^T - A \otimes I]\operatorname{vec}(X) = \operatorname{vec}(A^TX - XA^T),\\
[B \otimes I]\operatorname{vec}(X) = \operatorname{vec}(XB^T). \tag{2}
$$
where vec denotes the vectorization operator. Now, take the transpose of both sides of (1).  A matrix will have full column-rank if and only if its kernel is zero.  Combining this with (2) tells us that (1) will hold if and only if the linear map $\Phi:\Bbb R^{n \times n} \times \Bbb R^{n \times m} \to \Bbb R^{n \times n}$ defined by
$$
\Phi(X,Y) = A^TX - XA^T + B^TY
$$
has a trivial kernel.  That is, (1) will hold if and only if 
$$
A^TX - XA^T + YB^T = 0
$$
implies that $X = 0, Y = 0$ for all $X \in \Bbb R^{n \times n}$ and $Y \in \Bbb R^{n \times m}$.  Equivalently, this means that
$$
AX - XA = BY
$$
implies that $X = 0, Y = 0$.
This is turn equivalent to saying that for every non-zero matrix $X$, the column space 
$$
AX - XA
$$
cannot lie within the column space of $B$.
