If I have the following:
How do I show the following: $P_{11} = G_{11} + G_{12}\hat Y\tilde MG_{21}$ is:
I am stuck in this complicated system.
Or, the other simpler one: How do I show the following: $P_{12} = -G_{12}\hat M$ :
It seems the answer is not derived from just series combination of both.
Notice that G is a 2x2 system. The input/output group $ij$ enters via $B_i$ and goes out via $C_j$ hence $G_{ij}$ refers to $C_j(sI-A)^{-1}B_i$
Then,
$$ G_{11} = \begin{bmatrix}A&B_1\\C_1 &0\end{bmatrix}, G_{12} = \begin{bmatrix}A&B_2\\C_1 &D_{12}\end{bmatrix}, ... $$ and so on.
Thus, now if you do the tedious multiplication exercise (which I have to omit) you'll get the answer.Two state space systems admit multiplication $G_2G_1$ if output number of $G_1$ is equal to the input number of $G_1$. Then by substituting the $y$ equation of $G_2$ and rearranging terms give
$$ G_2G_1 = \begin{bmatrix}A_2&B_2C_1&B_2D_1\\0 &A_1&B_1\\C_2&D_2C_1&D_2D_1\end{bmatrix} $$
Thus we have, $$ \begin{align} P_{12} = G_{12}\hat M &=\begin{bmatrix}A&B_2\\C_1 &D_{12}\end{bmatrix} \begin{bmatrix}A+B_2K&B_2\\K &I\end{bmatrix}\\& = \begin{bmatrix}A&B_2K&B_2\\0&A+B_2K&B_2\\C_1&D_{12}K&D_{12}\end{bmatrix}\\ &= \begin{bmatrix}A_p&B_p\\C_p &D_{p}\end{bmatrix} \end{align} $$
The remaining trick is to see that the last expression has uncontrollable states and can be reduced. Let $$ T=\begin{bmatrix}I&0\\I&I \end{bmatrix} $$ Then if we do the state transformation, $\hat x=Tx$ $$ P_{12} = \begin{bmatrix}A_p&B_p\\C_p &D_{p}\end{bmatrix} = \begin{bmatrix}T^{-1}A_pT&T^{-1}B_p\\C_pT &D_{p}\end{bmatrix} = \begin{bmatrix}A+B_2K&B_2K&B_2\\0&A+B_2K&0\\C_1+D_{12}K&D_{12}K&D_{12}\end{bmatrix} $$ we see that the second subsystem is uncontrollable and can be removed and the minimal realization is as given in the problem.