Full rank lifted system description of state-space (A,B,C,D) For a state-space realization $(A,B,C,D)$, with $A\in \mathbb{R}^{n_x \times n_x}$, $B \in \mathbb{R}^{n_x \times n_u}$, $C \in \mathbb{R}^{n_y \times n_x}$, and  $D \in \mathbb{R}^{n_x \times n_u}$ I want to determine when the following matrix is full rank
\begin{equation}
\text{rank}
\begin{bmatrix}
I_{n_x\times n_x}-A^N & \mathcal{C}^e\\ 
\mathcal{O}^e & -J
\end{bmatrix}, \;\;\;\;\;\;\;\;\;\;\;(1)
\end{equation}
with $\mathcal{C}^e$ the extended controllability matrix 
\begin{equation}
\mathcal{C}^e= \begin{bmatrix}A^{N-1}B &  \cdots & AB & B
\end{bmatrix},
\end{equation}
$\mathcal{O}^e$ the extended observability matrix 
\begin{equation}
\mathcal{O}^e=\begin{bmatrix}C\\CA \\ \vdots \\ CA^{N-1}\end{bmatrix},
\end{equation}
and $J$ the lower triangular impulse response matrix 
\begin{equation}
J = \begin{bmatrix}
D & 0 & \cdots  & 0 \\ 
CB & D & \cdots & 0\\ 
\vdots & \vdots & \ddots & \vdots\\ 
CA^{N-2}B & \cdots & CB & D
\end{bmatrix}. \end{equation}
In MATLAB it turns out that this matrix (1) (for the cases I tested) is full rank $(N+n_x)$, in case $(A,B)$ is controllable and $(A,C)$ is observable, i.e. $(A,B,C,D)$ is a minimal realization. It should be noted that a stable (discrete-time) system is considered, i.e. $|\lambda(A)| < 1$ ($A$ Schur).
Via a Schur complement (1) can be written  as
\begin{equation}
\text{rank}
\begin{bmatrix}
J+\mathcal{O}^e(I_{n_x\times n_x}-A^N)^{-1}\mathcal{C}^e & 0_{N\times n_x}\\ 
0_{n_x\times N} & I_{n_x\times n_x}
\end{bmatrix} = 
n_x + \text{rank}\left(J+\mathcal{O}^e(I_{n_x\times n_x}-A^N)^{-1}\mathcal{C}^e\right).
\end{equation}
So now it reduces to proving that $\text{rank}\left(J+\mathcal{O}^e(I_{n_x\times n_x}-A^N)^{-1}\mathcal{C}^e\right)=N$. Can somebody help me proving this is true (for $(A,B,C,D)$ a minimal realization)? Or if it is easier to prove the rank of (1), that (1) is rank $N+n_x$? Note that for simplicity $n_y=n_u=1$ is chosen. Further it should be noted that the inverse of $(I_{n_x\times n_x}-A^N)$ always exists for $A$ Schur, since 
\begin{equation}
\nonumber
\text{det}(I-A^N) = \prod_i \lambda_i(I-A^N) = \prod_i\left(1-\lambda_i(A^N)\right) = \prod_i\left(1-\lambda^N_i(A)\right) \neq 0.
\end{equation} 
The rank of $J$ is typically $N-1$, since the direct feedthrough term is $D=0$ in most cases. The term $\mathcal{O}^e(I_{n_x\times n_x}-A^N)^{-1}\mathcal{C}^e$ has at most rank $n_x$ (Cayley-Hamilton theorem). However, combined this should be rank $N$.
Below an example of the matrix $(J+\mathcal{O}^e(I_{n_x\times n_x}-A^N)^{-1}\mathcal{C}^e)$ for $N=3$ is given
\begin{equation}
\nonumber
\begin{bmatrix}
D+C(I-A^3)^{-1}A^2B & C(I-A^3)^{-1}AB & C(I-A^3)^{-1}B \\ 
CB+C(I-A^3)^{-1}A^3B & D+C(I-A^3)^{-1}A^2B & C(I-A^3)^{-1}AB \\ 
CAB+C(I-A^3)^{-1}A^4B & CB+C(I-A^3)^{-1}A^3B & D+C(I-A^3)^{-1}A^2B 
\end{bmatrix},
\end{equation}
it can be seen that this matrix is Toeplitz.
 A: Assuming that the direct feedthrough term is zero ($D=0$), the matrix can be written as
\begin{equation}
\nonumber
\underbrace{\begin{bmatrix}
C\left(I-A^N\right)^{-1} &  & 0\\ 
 & \ddots & \\ 
0 &  & C\left(I-A^N\right)^{-1}
\end{bmatrix}}_{\mathbb{R}^{N\times Nn_x}}
\underbrace{\begin{bmatrix}
A^{N-1} & A^{N-2} & \cdots & I\\ 
I & A^{N-1} & \cdots & A\\ 
\vdots & \vdots & \ddots & \vdots\\ 
A^{N-2} & I  & \cdots & A^{N-1} 
\end{bmatrix}}_{\mathbb{R}^{Nn_x\times Nn_x}}
\underbrace{\begin{bmatrix}
B &  & 0\\ 
 & \ddots & \\ 
0 &  & B
\end{bmatrix}}_{\mathbb{R}^{Nn_x\times N}}
\end{equation}
Using Gaussian elimination, the middle matrix can be written as identity matrix for $A$ Schur, and thus this matrix is full rank. Then for $C$ full row rank and $B$ full column rank, the first and last matrix in the equation above have rank $N$.  Now combining different rank inequalities, it can be proved that the product of these three matrices is always full rank if: $A$ is Schur, the system is strictly proper ($D = 0$), and $(A,B,C)$ is a minimal realization, implying that pair $\{A,B\}$ is controllable, and pair $\{A,C\}$ is observable.
