Let $\hat M$ be am $m\times n$ matrix in $\mathbb{C}$ with $m\le n$. Let $\hat P=(\mathbb{1}|0)$ be another $m\times n$ matrix, projecting a $n$-dimensional space $V_1$ to a $m$-dimensional space $V_2$. $\hat P$ removes the last $m-n$ components of a vector in $V_1$. Let $\hat T$ and $\hat A$ be two square matrices such that
$$
\hat P=\hat A\hat P\hat T^{-1},
$$
that is, we have
$$
\hat T=\left(
\begin{array}{cc}
\hat A && \hat 0 \\
\hat B && \hat C
\end{array}
\right),
$$
Let us consider the following matrix transformation
$$
M\rightarrow \hat A\hat M\hat T^{-1}.
$$
We also allow additional column permutations so that the transformed projection $\hat P$ removes the $i_1,\dots i_{n-m}$-th components (instead of the last $m-n$ components).
If there is no vector $\vec x$ such that $\hat P\vec x\notin V_2$ and $\hat M\vec x=\vec 0$, then it is possible to reduce the matrix $\hat M$ to the form
$$
\left(
\begin{array}{c,c,c,c}
\hat J && \hat 0 && \dots && \hat 0 \\
\hat 0 && \hat D_1 &&\dots && \hat 0 \\
\vdots && \vdots && \ddots && \vdots \\
\hat 0 && \hat 0 && \dots && \hat D_{n-m}
\end{array}
\right)
$$
where $\hat J$ is a square matrix in Jordan form and $\hat D_k$ are single-block Jordan matrices with zero diagonal elements and the last row removed [they are $(s-1)\times s$ matrices].
In particular, if $m=n-1$ and $\hat M$ is a general matrix, it is possible to transform the matrix to the form
$$
\left(
\begin{array}{ccccc}
0 && 1 && 0 && \dots && 0 \\
0 && 0 && 1 &&\dots && 0 \\
\vdots && \vdots &&\vdots && \ddots && 0 \\
0 && 0 && 0 && \dots && 1
\end{array}
\right)
$$
apart from a zero-measure set of matrices.
I am wondering if this reduction is known in literature and if there is an obvious proof from the original proof of Jordan reduction for square matrices.
Let us give a numerical example.
$\hat M=\left(
\begin{array}{ccc}
9 & 2 & 1 \\
1 & 10 & 9 \\
\end{array}
\right)$, $\hat T=\left(
\begin{array}{ccc}
9 & -1 & 0 \\
80 & 8 & 0 \\
728 & 240 & 152 \\
\end{array}
\right)$, $\hat A=\left(
\begin{array}{cc}
9 & -1 \\
80 & 8 \\
\end{array}
\right)$.
We have that $\hat A \hat M\hat T^{-1}=\left(
\begin{array}{ccc}
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)$.
Someone could naively believe that this decomposition can be deduced by Jordan decomposition by adding a zero row to $\hat M$.
$M_1=\left(
\begin{array}{ccc}
9 & 2 & 1 \\
1 & 10 & 9 \\
0 & 0 & 0 \\
\end{array}
\right)$.
This can be transformed to the diagonal matrix
$T_1.M_1.T_1^{-1}=\left(
\begin{array}{ccc}
11 & 0 & 0 \\
0 & 8 & 0 \\
0 & 0 & 0 \\
\end{array}
\right)$ with
$\hat T_1=\left(
\begin{array}{ccc}
\frac{1}{3} & \frac{2}{3} & \frac{19}{33} \\
-\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\
0 & 0 & \frac{1}{11} \\
\end{array}
\right)$,
but the matrix $\hat T_1$ does not have the desired form of $\hat T$ with zeros in top-right block.
Best Answer
(The answer is completely replaced)
Consider together with $M$ another $m \times n$ matrix $N = [I|0]$. Transformations $M \mapsto AMT^{-1}$ with $T = \begin{bmatrix}A & 0 \\ B & C\end{bmatrix}$ are exactly the transformations of this from which preserve $N$, so the pair of matrices $(N, M)$ maps to $(N, \hat{M})$ under the transformation $(N, M) \mapsto (ANT^{-1}, AMT^{-1})$.
Now, from the Kronecker-Weierstrass theory of matrix pencils we know that $(N, M)$ can be transformed into a normal form, which is a direct sum of blocks of the forms $(I, J_\lambda)$, $(J_0, I)$ (where $J_\lambda$ is a Jordan block with eigenvalue $\lambda$), $([I|0],[0|I])$ of size $s \times (s+1)$, and $(\begin{bmatrix}I\\0\end{bmatrix},\begin{bmatrix}0\\I\end{bmatrix})$ of size $(s+1) \times s$.
Because $N$ is a full rank matrix, the second and the last types are not possible, so we transform $(N, M)$ into $(N', M')$ which is the sum of blocks $(I, J_\lambda)$ and $([I|0],[0|I])$. We can then apply a permutation of columns to restore $N$ and get $M'$ into the form you have in your post.