special relation between the expression of a matrix by block and its rank Given two matrices $A$ and $C$ of order $n\times n$ and $m\times n$ respectively. We define the following matrix by block:
$$
D=\left( \begin{matrix}
C\\
CA\\
:\\
:\\
CA^{q-1}
\end{matrix}\right)
$$
where $q\in\mathbb N^*$.
My question is: is there any relation between $q$ and the rank of the matrix $D$? if so for what value of $q$ we have $rank(D)=n$ ? Thank you for your time.
 A: We are given that $A\in M_{n\times n}$ and $C\in M_{m\times n}$. Note that $C A^\ell\in M_{m\times n}$ for any $\ell\geq 0$. It follows that $D\in M_{qm\times n}$. This gives the crude relationship between $\DeclareMathOperator{rank}{rank}\rank(D)$ and $q$
$$
\rank(D)\leq\min\{qm,n\}
$$
I don't think there's much hope to do any better than this without conditions on $A$ and $C$. For instance, if $C=I$ and $A$ is nilpotent with degree $k$ where $1\leq k\leq q-1$ then the rank of $D$ is a function of $k$.
A: First, it does not make sense to let $q\ge n$, since by Caley-Hamilton the powers $A^0 \dots A^n$ are linearly independent. Hence the rank of $D$ will not increase anymore for $q\ge n$.
The rank of $D$ is also limited by the properties of the matrices $A$ and $C$. If $A=0$ then $rank(D)=rank(C)$ no matter how large $q$ is.
The rank of $D$ is full ($rank(D)=n$) if the pair $(A,C)$ is observable. See http://en.wikipedia.org/wiki/Observability . The basic idea is that for observable systems, the solution of $\dot x=Ax$ can be reconstructed from $Cx(t)$.
