I am looking at the system \begin{eqnarray*} \begin{pmatrix} \dot{x}_1 \\ \dot{x}_2 \end{pmatrix} &=& \begin{pmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} + \begin{pmatrix} B_1 \\ B_2 \end{pmatrix} u,\\ y &=& \begin{pmatrix} I_{p} & 0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} \end{eqnarray*} which is assumed to be detectable. That is, there exists a matrix $\tilde{G}$ such that \begin{equation} \sigma \left( \begin{pmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{pmatrix} - \tilde{G} \begin{pmatrix} I_{p} & 0 \end{pmatrix} \right) \subset \mathbb C^{-}.\quad (*) \end{equation} I want to show that this implies that the pair $(A_{12},A_{22})$ is also detectable, that is, to show that there exists a matrix $G$ such that \begin{equation} \sigma(A_{22} - GA_{12}) \subset \mathbb C^{-}\quad (**) \end{equation} is also satisfied.
My working so far is as follows:
We know that $\tilde{G} \in \mathbb R^{n \times p }$ can be written as $\tilde{G} = \begin{pmatrix} G_1^T & G_2^T\end{pmatrix}^T$ so that the previous equation becomes \begin{equation} \sigma \begin{pmatrix} A_{11} - G_1& A_{12} \\ A_{21} -G_2 & A_{22} \end{pmatrix} \subset \mathbb C^{-}. \end{equation} I am unsure how to continue after this. I am think I am probably on the right track though. I want to deduce something about the eigenvalues in $(**)$ using the information of the eigenvalues in $(*)$.
Thanks in advance for any help!
An equivalent definition of detectability is the following:
Let $A,C$ matrices with dimensions $n\times n$ and $p\times n$ respectively. $(C,A)$ is detectable if $rank\left[\matrix{C\\A-\lambda \mathbb{I}}\right]=n$ for all $\lambda\in \mathbb{C}^+$.
From the detectability assumption we have that $$rank\left[\matrix{\mathbb{I}_p & 0\\A_{11}-\lambda \mathbb{I}_p & A_{12}\\ A_{21} & A_{22}-\lambda\mathbb{I}_{n-p}}\right]=n\:,\qquad \forall \lambda\in \mathbb{C}^+$$ In the above matrix the first $p$ columns are linearly independent from the rest $n-p$ columns. Thus, in order the matrix to have full column rank $n$ the following must hold $$rank\left[\matrix{ 0\\ A_{12}\\ A_{22}-\lambda\mathbb{I}_{n-p}}\right]=n-p\:,\qquad \forall \lambda\in \mathbb{C}^+$$ or equivalently $$rank\left[\matrix{ A_{12}\\ A_{22}-\lambda\mathbb{I}_{n-p}}\right]=n-p\:,\qquad \forall \lambda\in \mathbb{C}^+$$ The last condition means that $(A_{12},A_{22})$ is detectable.