[proofplan]
The coordinate change separates the measured component $z_1=y$ from the unmeasured component $z_2$ for the plant $\dot x=Ax+Bu$, $y=Cx$. We write the transformed plant equations in these coordinates, differentiate the error $e_2=z_2-w-My$, and substitute both the plant dynamics and the observer dynamics. The observer coefficients have been chosen so that every term involving the measured coordinate and the input cancels, leaving precisely the homogeneous error equation governed by $A_{22}-MA_{12}$. The observability hypothesis justifies the standard gain-selection premise that one may choose $M$ with $A_{22}-MA_{12}$ Hurwitz; after such an $M$ is fixed, the displayed error identity is purely algebraic and does not use observability again.
[/proofplan]
[step:Write the transformed plant equations in output-splitting coordinates]
Let $I\subset\mathbb{R}$ be the time interval on which the classical solution is defined. The statement declares the plant matrices $A\in\mathbb{R}^{n\times n}$, $B\in\mathbb{R}^{n\times m}$, and $C\in\mathbb{R}^{p\times n}$, together with the state map $x:I\to\mathbb{R}^n$, the input map $u:I\to\mathbb{R}^m$, the output map $y:I\to\mathbb{R}^p$, and the transformed state map $z:I\to\mathbb{R}^n$ defined by $z(t)=Tx(t)$. Write $z(t)=(z_1(t),z_2(t))$ with $z_1(t)\in\mathbb{R}^p$ and $z_2(t)\in\mathbb{R}^{n-p}$. The statement also declares the conformable block decomposition of $TAT^{-1}$ with upper-left block $A_{11}$, upper-right block $A_{12}$, lower-left block $A_{21}$, and lower-right block $A_{22}$, and the conformable block decomposition of $TB$ with upper block $B_1$ and lower block $B_2$. Since the plant and observer are being considered along classical trajectories, the maps $z_1:I\to\mathbb{R}^p$, $z_2:I\to\mathbb{R}^{n-p}$, and $w:I\to\mathbb{R}^{n-p}$ are differentiable on $I$.
Because $CT^{-1}$ is the block row matrix with first block $I_p$ and second block $0$, the output satisfies
\begin{align*}
y(t)=Cx(t)=CT^{-1}z(t)=z_1(t).
\end{align*}
Using the block decompositions of $TAT^{-1}$ and $TB$ declared in the statement, the transformed plant equation $\dot z=T\dot x=TAT^{-1}z+TBu$ is equivalent to
\begin{align*}
\dot z_1=A_{11}z_1+A_{12}z_2+B_1u.
\end{align*}
The second block is
\begin{align*}
\dot z_2=A_{21}z_1+A_{22}z_2+B_2u.
\end{align*}
Since $z_1=y$, these are also the equations for the measured coordinate $y$ and the unmeasured coordinate $z_2$.
[/step]
[step:Differentiate the reduced-order estimation error]
The reduced-order estimate is the map $\hat z_2:I\to\mathbb{R}^{n-p}$ defined by
\begin{align*}
\hat z_2=w+My.
\end{align*}
Since $y=z_1$, the error map $e_2:I\to\mathbb{R}^{n-p}$ is
\begin{align*}
e_2=z_2-\hat z_2=z_2-w-Mz_1.
\end{align*}
Differentiating this identity with respect to time gives
\begin{align*}
\dot e_2=\dot z_2-\dot w-M\dot z_1.
\end{align*}
Substitute the transformed plant equations and the observer equation:
\begin{align*}
\dot e_2=(A_{21}z_1+A_{22}z_2+B_2u)-\dot w-M(A_{11}z_1+A_{12}z_2+B_1u).
\end{align*}
[/step]
[step:Substitute the observer dynamics and cancel the measured and input terms]
Using $y=z_1$, the observer equation is
\begin{align*}
\dot w=(A_{22}-MA_{12})w+\bigl(A_{21}-MA_{11}+(A_{22}-MA_{12})M\bigr)z_1+(B_2-MB_1)u.
\end{align*}
Substituting this expression into the differentiated error equation yields
\begin{align*}
\dot e_2=A_{21}z_1+A_{22}z_2+B_2u-(A_{22}-MA_{12})w-\bigl(A_{21}-MA_{11}+(A_{22}-MA_{12})M\bigr)z_1-(B_2-MB_1)u-MA_{11}z_1-MA_{12}z_2-MB_1u.
\end{align*}
Collecting the input terms gives
\begin{align*}
B_2u-(B_2-MB_1)u-MB_1u=0.
\end{align*}
Collecting the measured-coordinate terms gives
\begin{align*}
A_{21}z_1-\bigl(A_{21}-MA_{11}+(A_{22}-MA_{12})M\bigr)z_1-MA_{11}z_1=-(A_{22}-MA_{12})Mz_1.
\end{align*}
The remaining $z_2$ and $w$ terms are therefore
\begin{align*}
\dot e_2=A_{22}z_2-MA_{12}z_2-(A_{22}-MA_{12})w-(A_{22}-MA_{12})Mz_1.
\end{align*}
Thus
\begin{align*}
\dot e_2=(A_{22}-MA_{12})z_2-(A_{22}-MA_{12})(w+Mz_1).
\end{align*}
[guided]
The transformed state map is $z:I\to\mathbb{R}^n$, $z(t)=Tx(t)$, and we write $z(t)=(z_1(t),z_2(t))$ with $z_1(t)\in\mathbb{R}^p$ and $z_2(t)\in\mathbb{R}^{n-p}$. Since $CT^{-1}=\begin{pmatrix}I_p&0\end{pmatrix}$, the output map $y:I\to\mathbb{R}^p$ satisfies $y=z_1$. The block decompositions of $TAT^{-1}$ and $TB$ give the transformed plant equations
\begin{align*}
\dot z_1=A_{11}z_1+A_{12}z_2+B_1u
\end{align*}
and
\begin{align*}
\dot z_2=A_{21}z_1+A_{22}z_2+B_2u.
\end{align*}
The purpose of the reduced-order observer is to estimate only $z_2$, because $z_1$ is already measured through $y=z_1$. The observer is therefore built around the corrected variable $w:I\to\mathbb{R}^{n-p}$, with the estimate recovered by $\hat z_2=w+My$. Since $y=z_1$, the error is
\begin{align*}
e_2=z_2-w-Mz_1.
\end{align*}
Differentiating gives
\begin{align*}
\dot e_2=\dot z_2-\dot w-M\dot z_1.
\end{align*}
Now use the transformed plant equations
\begin{align*}
\dot z_1=A_{11}z_1+A_{12}z_2+B_1u
\end{align*}
and
\begin{align*}
\dot z_2=A_{21}z_1+A_{22}z_2+B_2u.
\end{align*}
The observer equation, after replacing $y$ by $z_1$, is
\begin{align*}
\dot w=(A_{22}-MA_{12})w+\bigl(A_{21}-MA_{11}+(A_{22}-MA_{12})M\bigr)z_1+(B_2-MB_1)u.
\end{align*}
Substitution gives
\begin{align*}
\dot e_2=A_{21}z_1+A_{22}z_2+B_2u-(A_{22}-MA_{12})w-\bigl(A_{21}-MA_{11}+(A_{22}-MA_{12})M\bigr)z_1-(B_2-MB_1)u-MA_{11}z_1-MA_{12}z_2-MB_1u.
\end{align*}
We now group terms by the signal they multiply. The input contribution is
\begin{align*}
B_2u-(B_2-MB_1)u-MB_1u=0.
\end{align*}
This is exactly why the observer input coefficient is $B_2-MB_1$: it cancels the direct input contribution to the error dynamics.
Next collect the terms multiplying the measured coordinate $z_1$. They are
\begin{align*}
A_{21}z_1-\bigl(A_{21}-MA_{11}+(A_{22}-MA_{12})M\bigr)z_1-MA_{11}z_1.
\end{align*}
Expanding the middle term and cancelling $A_{21}z_1$ with $-A_{21}z_1$ and $MA_{11}z_1$ with $-MA_{11}z_1$ leaves
\begin{align*}
-(A_{22}-MA_{12})Mz_1.
\end{align*}
This is exactly why the measured-coordinate coefficient contains $A_{21}-MA_{11}+(A_{22}-MA_{12})M$.
After these cancellations, the remaining terms are
\begin{align*}
\dot e_2=A_{22}z_2-MA_{12}z_2-(A_{22}-MA_{12})w-(A_{22}-MA_{12})Mz_1.
\end{align*}
Factoring the common matrix $A_{22}-MA_{12}$ gives
\begin{align*}
\dot e_2=(A_{22}-MA_{12})z_2-(A_{22}-MA_{12})(w+Mz_1).
\end{align*}
[/guided]
[/step]
[step:Factor the remaining expression through the error]
Since $\hat z_2=w+My=w+Mz_1$, the preceding identity becomes
\begin{align*}
\dot e_2=(A_{22}-MA_{12})(z_2-\hat z_2).
\end{align*}
By the definition $e_2=z_2-\hat z_2$, this is
\begin{align*}
\dot e_2=(A_{22}-MA_{12})e_2.
\end{align*}
This proves the asserted reduced-order error equation.
[/step]
