rank of this matrix

Question

• ${\rm rank\,}\begin{bmatrix}B&AB&\cdots&A^{n-1}B\end{bmatrix}=n$ 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

Answer 1

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 $B$.