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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.

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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 $$\operatorname{rank}\begin{bmatrix}A_2-\lambda I & B_2 T_1(\lambda)\end{bmatrix}=n_2, ~~ \forall \lambda \in \sigma(A_2)$$

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  • $\begingroup$ One follow up question: in your third sentence you state that to split up the matrix C we require that the spectrum of A1 and A2 are distinct. Is this not equivalent to what is stated near the beginning of the question? More specifically; the part written as ' σ(A1)∩σ(A2)=∅'. $\endgroup$ – piece_and_love Dec 26 '18 at 20:33
  • $\begingroup$ Yes but you need to mention it in your proof, like "because of this assumption we can do this", etc. So the reader can understand that it only applies for this specific assumption and not true in general. If your assumption does not make any difference then you should write "without losing generality, we can assume that...". $\endgroup$ – obareey Dec 27 '18 at 14:50
  • $\begingroup$ Sorry to bother you after such a long time; is it required that $w_1$ and $w_2$ are eigenvectors of $A_1$ and $A_2$ respectively? $\endgroup$ – piece_and_love Jan 2 at 23:25
  • $\begingroup$ It should be true for all vectors, but checking only for the eigenvectors is enough as $w^T(A-\lambda I)=0$ if and only if $w$ is a left eigenvector of $A$ for the eigenvalue $\lambda$, by definition. $\endgroup$ – obareey Jan 3 at 7:49
  • $\begingroup$ And so to be absolutely clear; $w_1^T(A_1-\lambda I ,\,\, B_1)=0 \iff w_1^T=0$ is equivalent to $rank[A_1-\lambda I ,\,\, B_1]=n_1$? I understand that it's a necessary condition for full rank but is it alone sufficient? $\endgroup$ – piece_and_love Jan 3 at 17:08

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