We know that for an uncontrollable state space model of the following form:
$\begin{array}{l} \dot x\left( t \right) = Ax\left( t \right) + Bu\left( t \right)\\ y\left( t \right) = Cx\left( t \right) + Du\left( t \right) \end{array}$
it has a transfer function $G\left( s \right) = C{\left( {SI - A} \right)^{ - 1}}B + D$.
Furthermore, we can associate a controllable Kalman decomposition:
$\begin{array}{l} \left[ {\begin{array}{*{20}{c}} {\frac{{d\left( {{{\bar x}_c}} \right)}}{{dt}}}\\ {\frac{{d\left( {{{\bar x}_{\bar c}}} \right)}}{{dt}}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {{{\bar A}_c}}&{{{\bar A}_{12}}}\\ 0&{{{\bar A}_{\bar c}}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{{\bar x}_c}}\\ {{{\bar x}_{\bar c}}} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} {{{\bar B}_c}}\\ 0 \end{array}} \right]u\\ y = \left[ {\begin{array}{*{20}{c}} {{{\bar C}_c}}&{{{\bar C}_{\bar c}}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{{\bar x}_c}}\\ {{{\bar x}_{\bar c}}} \end{array}} \right] + Du \end{array}$
my question is: how to prove that the new transfer function is: $G\left( s \right) = {{\bar C}_c}{\left( {SI - {{\bar A}_c}} \right)^{ - 1}}{{\bar B}_c} + D$
The controllable Kalman decomposition is a linear transformation, i.e. $\bar{x} = T\,x$ with $T$ non-singular. So the state space model in the controllable Kalman decomposition $(\bar{A}, \bar{B}, \bar{C}, \bar{D})$ is related to the initial state space model $(A,B,C,D)$ according to
\begin{align} \bar{A} &= T\,A\,T^{-1}, \\ \bar{B} &= T\,B, \\ \bar{C} &= C\,T^{-1}, \\ \bar{D} &= D. \end{align}
For any such non-singular $T$ is follows that
$$ G(s) = \bar{C}\,(s\,I - \bar{A})^{-1} \bar{B} + \bar{D} = C\,T^{-1}\,(s\,I - T\,A\,T^{-1})^{-1} T\,B + D, \tag{1} $$
which simplifies to the following when using the push-through identity
$$ G(s) = C\,(s\,I - A)^{-1} B + D. \tag{2} $$
So any non-singular $T$ does not change the transfer function $G(s)$ associated with the state space model.
If there exists such a $T$ such that
$$ \bar{A} = \begin{bmatrix} \bar{A}_c & \bar{A}_{12} \\ 0 & \bar{A}_\bar{c} \end{bmatrix}, \quad \bar{B} = \begin{bmatrix} \bar{B}_c \\ 0 \end{bmatrix}, \quad \bar{C} = \begin{bmatrix} \bar{C}_c & \bar{C}_\bar{c} \end{bmatrix}, $$
it follows that the transfer function $G(s)$ can be obtained using $(1)$. It can be noted that due to upper block triangular structure of $\bar{A}$ it holds that
\begin{align} (s\,I - \bar{A})^{-1} &= \begin{bmatrix} s\,I-\bar{A}_c & -\bar{A}_{12} \\ 0 & s\,I-\bar{A}_\bar{c} \end{bmatrix}^{-1}, \tag{3a} \\ &= \begin{bmatrix} (s\,I-\bar{A}_c)^{-1} & (s\,I-\bar{A}_c)^{-1} \bar{A}_{12} (s\,I-\bar{A}_\bar{c})^{-1} \\ 0 & (s\,I-\bar{A}_\bar{c})^{-1} \end{bmatrix}. \tag{3b} \end{align}
From this it follows that multiplying $(3)$ by $\bar{B}$ yields
$$ (s\,I - \bar{A})^{-1} \bar{B} = \begin{bmatrix} (s\,I-\bar{A}_c)^{-1} \bar{B}_c \\ 0 \end{bmatrix}, $$
after which I assume it is clear that
$$ G(s) = \bar{C}_c(s\,I-\bar{A}_c)^{-1} \bar{B}_c + D. $$