I have two linear control systems that are represented by their state space models
$$\left(
\begin{array}{c|c}
A_1 & B_1 \\
\hline
C_1 & D_1 \\
\end{array}
\right),
\left(
\begin{array}{c|c}
A_2 & B_2 \\
\hline
C_2 & D_2 \\
\end{array}
\right)$$
where $A_i$ is the state matrix, $B_i$ is the input matrix, $C_i$ is the output matrix and $D_i$ is the feed-forward matrix. With $\sigma(A_1)\cap\sigma(A_2)=\emptyset$.
The output of the first system is a vector signal of dimension $n_1$, which is the same dimension of the input signal of the second system.
I have found that the state space representation is given by:
$$\Sigma_G: \begin{cases}
x'(t) = Ax(t)+Bu(t)\\
y(t)=Cx(t)+Du(t)
\end{cases}$$
where $A=\left(
\begin{array}{cc}
A_1 & 0 \\
B_2C_1 & A_2 \\
\end{array}
\right)$, $B=\begin{bmatrix} B_1\\B_2D_1 \end{bmatrix}$, $C=[D_2C_1\,\,\,\, C_2]$ and $D=[D_2D_1]$.
From all of this, I would like to show that $(A,B)$ is controllable if and only if $(A_1,B_1)$ is controllable and $rank(A_2-\lambda I \,\,\,\,\,\,B_2T_1(\lambda))=n_2$ for all $\lambda \in \sigma(A_2)$, where $n_2$ is the dimension of the second original state space and $T_1(s)$ is the transfer matrix of the first original state space.
My attempts/progress:
It is well documented that this transformation matrix, $T_1(\lambda)$, is given by $T_1(\lambda)=C_1 (\lambda I-A_1)^{-1}B_1+D_1$.
For $(A,B)$ to be controllable we could first find $C=rank[B\,\,\,AB\,\,\,A^2B … A^{n-1}B]$ though I'm not sure what value we should equate this to in order to show controllability. In finding $C$ I had the following workings:
$$AB=\begin{bmatrix} A_1B_1\\B_2C_1B_1+A_2B_2D_1 \end{bmatrix}$$ and $$A^2B=\begin{bmatrix} A_1^2B_1\\ B_2C_1A_1B_1+A_2(B_2C_1B_1+A_2B_2D_1) \end{bmatrix}.$$ Pretty soon I was able to see that the top part of the block Matrix C had the pattern of $$B_1, A_1B_1, A_1^2B_1,…$$ which resembles the condition of $(A_1,B_1)$ being controllable if and only if $$rank[B_1,A_1B_1,A_1^2B_1,…,A_1^{n-1}B_1]=n_1.$$
As for the bottom part of C, I see absolutely no pattern and so I have no idea how to approach the second part of the equivalence.
Is there a connection between solving this question and my observations?
After many wasted hours of failed attempts to solve this, I appreciate any help offered to me.
Best Answer
This is not a complete answer, but I think it can give some insight to the problem.
You cannot separate the top and bottom of the $C$ matrix in general. You have to use the assumption that the spectrum of $A_1$ and $A_2$ are distinct to do so.
We can prove the statement more easily with the Hautus criteria, which is equivalent to $$\begin{bmatrix}w_1^T && w_2^T\end{bmatrix}\begin{bmatrix}A_1-\lambda I && 0 && B_1 \\ B_2 C_1 && A_2-\lambda I && B_2 D_1\end{bmatrix} = 0 \iff \begin{bmatrix}w_1^T && w_2^T\end{bmatrix} = 0, ~~ \forall \lambda \in \sigma(A) \tag{1}$$ if and only if $(A,B)$ is controllable.
First, assume that $(A,B)$ is controllable. Then $(1)$ applies. Since the spectrum of $A_1$ and $A_2$ are distinct and $\sigma(A)=\sigma(A_1)\cup\sigma(A_2)$, we can first check for the $\lambda \in \sigma(A_1)$. Since $\lambda \notin \sigma(A_2)$, for any nonzero $w_2$, $w_2^T (A_2-\lambda I) \neq 0$, which means $(1)$ is satisfied. Now select $w_2=0$, which means $$w_1^T \begin{bmatrix}A_1-\lambda I && B_1\end{bmatrix} = 0 \iff w_1^T = 0, ~~ \forall \lambda \in \sigma(A_1)$$ So, $(A_1,B_1)$ is controllable.
Similarly, for $\lambda\in\sigma(A_2)$ and any nonzero $w_1$, $w_1^T(A_1-\lambda I)\neq0$, hence $(1)$ is satisfied. For $w_1=0$ we need $$w_2^T \begin{bmatrix}A_2-\lambda I && B_2 C_1 & B_2 D_1\end{bmatrix} = 0 \iff w_2^T = 0, ~~ \forall \lambda \in \sigma(A_2)$$
Now, we need to show somehow this is equivalent to $$\operatorname{rank}\begin{bmatrix}A_2-\lambda I & B_2 T_1(\lambda)\end{bmatrix}=n_2, ~~ \forall \lambda \in \sigma(A_2)$$