[proofplan]
We compute the transfer function in Kalman block coordinates. For a fixed spectral parameter $s$ and input vector $v$, we solve the resolvent equation $(sI-\tilde A)z=\tilde Bv$ block by block. The unreachable blocks are not forced by the input, and the reachable-unobservable block is annihilated by the output map. Substituting the resulting block solution into $\tilde C z+Dv$ leaves only the reachable-observable contribution.
[/proofplan]
[step:Pass to Kalman block coordinates without changing the transfer function]
Let $\tilde X:=X_{ro}\oplus X_{ru}\oplus X_{uo}\oplus X_{uu}$, and let $T:X \to \tilde X$ be the invertible coordinate map implementing the chosen Kalman decomposition. By definition of the transformed realization, $\tilde A = TAT^{-1}$, $\tilde B = TB$, $\tilde C = CT^{-1}$, and $\tilde D = D$. Let $\rho(A)$ denote the resolvent set of $A$, namely the set of scalars $s$ for which $sI_X-A:X\to X$ is invertible.
For every $s\in\rho(A)$, the operator $sI_X-A$ is invertible. Since
\begin{align*}
sI_{\tilde X}-\tilde A=T(sI_X-A)T^{-1},
\end{align*}
the operator $sI_{\tilde X}-\tilde A$ is invertible with inverse
\begin{align*}
(sI_{\tilde X}-\tilde A)^{-1}=T(sI_X-A)^{-1}T^{-1}.
\end{align*}
Indeed, multiplying $T(sI_X-A)T^{-1}$ by $T(sI_X-A)^{-1}T^{-1}$ on either side gives the identity operator on $\tilde X$.
Hence
\begin{align*}
\tilde C(sI_{\tilde X}-\tilde A)^{-1}\tilde B+\tilde D=C(sI_X-A)^{-1}B+D.
\end{align*}
It is therefore enough to compute the transfer function in the displayed Kalman coordinates.
[guided]
The transfer function is invariant under a change of state coordinates. Let $\tilde X:=X_{ro}\oplus X_{ru}\oplus X_{uo}\oplus X_{uu}$. The coordinate map $T:X \to \tilde X$ is an isomorphism, so the transformed realization is given by $\tilde A = TAT^{-1}$, $\tilde B = TB$, $\tilde C = CT^{-1}$, and $\tilde D = D$. Let $\rho(A)$ denote the resolvent set of $A$, namely the set of scalars $s$ for which $sI_X-A:X\to X$ is invertible.
If $s\in\rho(A)$, then $sI_X-A$ is invertible. Conjugating by $T$ shows that
\begin{align*}
sI_{\tilde X}-\tilde A=T(sI_X-A)T^{-1},
\end{align*}
so $sI_{\tilde X}-\tilde A$ is also invertible, with inverse
\begin{align*}
(sI_{\tilde X}-\tilde A)^{-1}=T(sI_X-A)^{-1}T^{-1}.
\end{align*}
Substituting this identity into the transformed transfer function gives
\begin{align*}
\tilde C(sI_{\tilde X}-\tilde A)^{-1}\tilde B+\tilde D=CT^{-1}T(sI_X-A)^{-1}T^{-1}TB+D.
\end{align*}
Cancelling $T^{-1}T=I_X$ and $TT^{-1}=I_{\tilde X}$ yields
\begin{align*}
\tilde C(sI_{\tilde X}-\tilde A)^{-1}\tilde B+\tilde D=C(sI_X-A)^{-1}B+D.
\end{align*}
Thus computing the transfer function in Kalman coordinates loses no information; it only exposes which state components can actually contribute to input-output behavior.
[/guided]
[/step]
[step:Force the unreachable coordinates to vanish from the triangular equations]
Fix $s$ in the common resolvent set of $\tilde A$ and $A_{ro}$, and fix an input vector $v\in U$, where $U$ is the input space. Define $z\in\tilde X$ by
\begin{align*}
z=(sI_{\tilde X}-\tilde A)^{-1}\tilde Bv.
\end{align*}
Write
\begin{align*}
z=z_{ro}\oplus z_{ru}\oplus z_{uo}\oplus z_{uu},
\end{align*}
with $z_{ro}\in X_{ro}$, $z_{ru}\in X_{ru}$, $z_{uo}\in X_{uo}$, and $z_{uu}\in X_{uu}$. The Kalman block form means that there are maps $B_{ro}:U\to X_{ro}$, $B_{ru}:U\to X_{ru}$, $A_{uo}:X_{uo}\to X_{uo}$, $A_{uu}:X_{uu}\to X_{uu}$, and $A_{43}:X_{uo}\to X_{uu}$ such that the $X_{uo}$ and $X_{uu}$ components of $\tilde Bv$ are zero and the relevant components of $\tilde A z$ are $A_{uo}z_{uo}$ and $A_{43}z_{uo}+A_{uu}z_{uu}$, respectively. Hence the $X_{uo}$ component of $(sI_{\tilde X}-\tilde A)z=\tilde Bv$ is
\begin{align*}
(sI_{X_{uo}}-A_{uo})z_{uo}=0.
\end{align*}
The $X_{uu}$ component is
\begin{align*}
-A_{43}z_{uo}+(sI_{X_{uu}}-A_{uu})z_{uu}=0.
\end{align*}
We use the following elementary consequence of finite-dimensional block triangular form.
[claim:Diagonal blocks of an invertible triangular operator are invertible]
Let $Y=Y_1\oplus\cdots\oplus Y_m$ be a finite-dimensional direct sum, and let $L:Y\to Y$ be block triangular with diagonal blocks $L_{jj}:Y_j\to Y_j$. If $L$ is invertible, then every $L_{jj}$ is invertible.
[/claim]
[proof]
Choose ordered bases of $Y_1,\dots,Y_m$ and concatenate them to obtain an ordered basis of $Y$. In that basis the matrix of $L$ is block triangular, with diagonal blocks represented by the matrices of $L_{jj}:Y_j\to Y_j$. Expanding the determinant by the Leibniz formula, any permutation that selects an entry from the triangular zero region contributes a zero product. The only nonzero products are therefore obtained by choosing entries within the diagonal blocks, so
\begin{align*}
\det L=\prod_{j=1}^{m}\det L_{jj}.
\end{align*}
Since $L$ is invertible, $\det L\ne 0$, so each factor $\det L_{jj}$ is nonzero. Since each $Y_j$ is finite-dimensional, nonzero determinant is equivalent to invertibility of $L_{jj}$.
[/proof]
Apply the claim to $L=sI_{\tilde X}-\tilde A$. Thus the diagonal blocks $sI_{X_{ro}}-A_{ro}$, $sI_{X_{ru}}-A_{ru}$, $sI_{X_{uo}}-A_{uo}$, and $sI_{X_{uu}}-A_{uu}$ are invertible. Hence the first displayed equation gives $z_{uo}=0$, and then the second displayed equation gives $z_{uu}=0$. The $X_{ro}$ block equation is therefore
\begin{align*}
(sI_{X_{ro}}-A_{ro})z_{ro}=B_{ro}v.
\end{align*}
Since $sI_{X_{ro}}-A_{ro}$ is invertible by the choice of $s$, we obtain
\begin{align*}
z_{ro}=(sI_{X_{ro}}-A_{ro})^{-1}B_{ro}v.
\end{align*}
[guided]
Fix $s$ in the common resolvent set of $\tilde A$ and $A_{ro}$, and fix an arbitrary input vector $v\in U$, where $U$ is the input space. Define $z\in\tilde X$ by
\begin{align*}
z=(sI_{\tilde X}-\tilde A)^{-1}\tilde Bv.
\end{align*}
Write $z=z_{ro}\oplus z_{ru}\oplus z_{uo}\oplus z_{uu}$ with $z_{ro}\in X_{ro}$, $z_{ru}\in X_{ru}$, $z_{uo}\in X_{uo}$, and $z_{uu}\in X_{uu}$.
The delicate point is not that an arbitrary diagonal block of an invertible block matrix is invertible; that statement is false in general. What we use is the specific triangular structure supplied by the Kalman decomposition. In these coordinates, the input block decomposes through maps $B_{ro}:U\to X_{ro}$ and $B_{ru}:U\to X_{ru}$, while the $X_{uo}$ and $X_{uu}$ components of $\tilde Bv$ are zero. The relevant Kalman blocks are maps $A_{uo}:X_{uo}\to X_{uo}$, $A_{uu}:X_{uu}\to X_{uu}$, and $A_{43}:X_{uo}\to X_{uu}$, and the $X_{uo}$ and $X_{uu}$ components of $\tilde A z$ are $A_{uo}z_{uo}$ and $A_{43}z_{uo}+A_{uu}z_{uu}$, respectively.
The $X_{uo}$ component of $(sI_{\tilde X}-\tilde A)z=\tilde Bv$ is
\begin{align*}
(sI_{X_{uo}}-A_{uo})z_{uo}=0.
\end{align*}
For this triangular operator, the determinant expansion uses only products of entries from diagonal blocks: any permutation selecting an entry from the triangular zero region contributes zero. Therefore
\begin{align*}
\det(sI_{\tilde X}-\tilde A)=\det(sI_{X_{ro}}-A_{ro})\det(sI_{X_{ru}}-A_{ru})\det(sI_{X_{uo}}-A_{uo})\det(sI_{X_{uu}}-A_{uu}).
\end{align*}
Since $sI_{\tilde X}-\tilde A$ is invertible, the determinant on the left is nonzero, so each factor on the right is nonzero. In finite dimension, a square operator has nonzero determinant exactly when it is invertible. Hence $sI_{X_{uo}}-A_{uo}$ is invertible, and the displayed $X_{uo}$ equation gives $z_{uo}=0$.
Now the $X_{uu}$ component is
\begin{align*}
-A_{43}z_{uo}+(sI_{X_{uu}}-A_{uu})z_{uu}=0.
\end{align*}
Substituting $z_{uo}=0$ gives
\begin{align*}
(sI_{X_{uu}}-A_{uu})z_{uu}=0.
\end{align*}
The same determinant argument shows that $sI_{X_{uu}}-A_{uu}$ is invertible, so $z_{uu}=0$. This is the exact reason the observable but unreachable block cannot contribute to the zero-state output: no input reaches it, and the resolvent equation forces its component to vanish.
With $z_{uo}=0$, the $X_{ro}$ component of the resolvent equation reduces to
\begin{align*}
(sI_{X_{ro}}-A_{ro})z_{ro}=B_{ro}v.
\end{align*}
Because $s$ was chosen in the resolvent set of $A_{ro}$, the operator $sI_{X_{ro}}-A_{ro}$ is invertible. Therefore
\begin{align*}
z_{ro}=(sI_{X_{ro}}-A_{ro})^{-1}B_{ro}v.
\end{align*}
[/guided]
[/step]
[step:Apply the output map and eliminate the hidden blocks]
Using the block form of $\tilde C$, there are output maps $C_{ro}:X_{ro}\to Y$ and $C_{uo}:X_{uo}\to Y$, while the $X_{ru}$ and $X_{uu}$ columns of $\tilde C$ are zero. Thus the transformed output corresponding to the input vector $v$ is
\begin{align*}
\tilde C z + Dv=C_{ro}z_{ro}+C_{uo}z_{uo}+Dv.
\end{align*}
From the previous step, $z_{uo}=0$ and
\begin{align*}
z_{ro}=(sI_{X_{ro}}-A_{ro})^{-1}B_{ro}v.
\end{align*}
Therefore
\begin{align*}
\tilde C(sI_{\tilde X}-\tilde A)^{-1}\tilde Bv+Dv=C_{ro}(sI_{X_{ro}}-A_{ro})^{-1}B_{ro}v+Dv.
\end{align*}
Because $v\in U$ was arbitrary, this proves the operator identity
\begin{align*}
\tilde C(sI_{\tilde X}-\tilde A)^{-1}\tilde B+D=C_{ro}(sI_{X_{ro}}-A_{ro})^{-1}B_{ro}+D.
\end{align*}
By the coordinate-invariance established in the first step, the same identity holds for $G(s)$.
[/step]
[step:Extend the equality as an identity of rational matrices]
Both sides are rational matrix-valued functions of $s$. The preceding computation proves equality for every $s$ in the common resolvent set on which the displayed inverses exist. For each matrix entry, the difference between the two sides is a rational function. If the state spaces are complex, its numerator polynomial vanishes on a nonempty open subset of $\mathbb{C}$, so it is the zero polynomial. If the state spaces are real, the same numerator polynomial vanishes on the infinite real subset of the common resolvent set obtained above, so it is again the zero polynomial. Therefore the difference is the zero rational function entrywise, the two rational matrix functions agree, and we obtain
\begin{align*}
G(s)=C_{ro}(sI_{X_{ro}}-A_{ro})^{-1}B_{ro}+D.
\end{align*}
The reachable-unobservable component may be forced through $B_{ru}$, but it is killed by the zero $X_{ru}$-column of $\tilde C$. The unreachable-observable component appears in $\tilde C$ through $C_{uo}$, but its resolvent component is not forced by $\tilde B$ from zero initial state. The unreachable-unobservable component is neither forced by the input nor seen by the output. Hence none of the three hidden diagonal modes changes the zero-state input-output map.
[guided]
Both sides are rational matrix-valued functions of the spectral parameter $s$. The previous steps proved equality at every $s$ for which both resolvents used in the computation exist. To pass from pointwise equality on that resolvent set to equality as rational matrices, examine one matrix entry at a time. The entrywise difference has the form $p(s)/q(s)$ for polynomials $p$ and $q$, with $q$ not the zero polynomial. On the common resolvent set, this rational function is zero, so $p(s)=0$ at every such value of $s$.
If the realization is over $\mathbb{C}$, the common resolvent set contains a nonempty open subset of $\mathbb{C}$ away from finitely many eigenvalues, and a polynomial vanishing on a nonempty open subset of $\mathbb{C}$ is the zero polynomial. If the realization is over $\mathbb{R}$, the common real resolvent set contains infinitely many real values because only finitely many real spectral values are excluded, and a real polynomial with infinitely many real zeros is the zero polynomial. In either case $p=0$, so the entrywise difference is the zero rational function.
Therefore the rational matrix identity is
\begin{align*}
G(s)=C_{ro}(sI_{X_{ro}}-A_{ro})^{-1}B_{ro}+D.
\end{align*}
The three hidden diagonal modes are excluded for distinct structural reasons: the reachable-unobservable component may be forced through $B_{ru}$ but is killed by the zero $X_{ru}$-column of $\tilde C$; the unreachable-observable component appears in $\tilde C$ through $C_{uo}$ but has zero resolvent component from zero initial state; and the unreachable-unobservable component is neither forced by the input nor observed by the output. Thus none of these modes changes the zero-state input-output map.
[/guided]
[/step]
