Writing a semi explicit ADE in the form of an implicit ODE My text says that we can take an ADE of the form
$$
\textbf{x}'=\textbf{f}(t,\textbf{x},\textbf{z})\\
\textbf{0}=\textbf{g}(t,\textbf{x},\textbf{z})
$$
and rewrite it as an ADE in the form
$$
\textbf{F}(t,\textbf{y},\textbf{y}')=0,
$$
where
$$
\textbf{y}=\begin{pmatrix}\textbf{x}\\\textbf{z}\end{pmatrix},
$$
and
$$
\frac{\partial\textbf{F}(t,\textbf{u},\textbf{v})}{\partial\textbf{v}}=\begin{pmatrix}I&0\\0&0\end{pmatrix},
$$
which is singular. My question is, how? They simply say that this is obvious.
What I figured was, writing $\textbf{x}=\textbf{y}_1$ and $\textbf{z}=\textbf{y}_2$, I could take 
$$
\textbf{F}(t,\textbf{y},\textbf{y}')=\begin{pmatrix}\textbf{y}_1'-\textbf{f}(t,\textbf{y})\\\textbf{g}(t,\textbf{y})\end{pmatrix}.
$$
Then if we say that $\textbf{x}\in\mathbb{R}^n$ and $\textbf{z}\in\mathbb{R}^m$, then $\textbf{y}'\in\mathbb{R}^{n+m}$ (note: I'm not given that $\textbf{z}$ is the same dimension as $\textbf{x}$). So I'm differentiating with respect to $n+m$ variables when I calculate the Jacobian, and their wording ("no longer nonsingular") suggests that I should end up with a square matrix somehow. So I would need $\textbf{F}\in\mathbb{R}^{n+m}$, I think. But it seems like I've suddenly made an assumption on the dimension of $\textbf{g}$, when no dimension was given.
So basically I don't know where to go from here or if I'm even on the right track. Help appreciated.
 A: Given $\mathbf{x} \in \mathbb{R}^n$ and $\mathbf{z}  \in \mathbb{R}^m$ as well as $\mathbf{f} :  \mathbb{R} \times \mathbb{R}^n \times \mathbb{R}^m  \to  \mathbb{R}^n$ and  $\mathbf{g} :  \mathbb{R} \times \mathbb{R}^n \times \mathbb{R}^m  \to  \mathbb{R}^m$. Simply differentiate the second equation with respect to time, assuming the variables are all time dependnet:
\begin{align}0 &=\frac{d}{dt} \mathbf{g} \big( t, \mathbf{x}(t), \mathbf{z}(t)\big) = \frac{\partial \mathbf{g}}{\partial t}( t, \mathbf{x}, \mathbf{z}) + \frac{\partial \mathbf{g}}{\partial \mathbf{x}}( t, \mathbf{x}, \mathbf{z})\cdot \mathbf{x'} + \frac{\partial \mathbf{g}}{\partial \mathbf{z}}( t, \mathbf{x}, \mathbf{z}) \cdot\mathbf{z}'\\
&= \frac{\partial \mathbf{g}}{\partial t}( t, \mathbf{x}, \mathbf{z}) + \frac{\partial \mathbf{g}}{\partial \mathbf{x}}( t, \mathbf{x}, \mathbf{z})\cdot \mathbf{f}( t, \mathbf{x}, \mathbf{z})   + \frac{\partial \mathbf{g}}{\partial \mathbf{z}}( t, \mathbf{x}, \mathbf{z}) \cdot\mathbf{z}'
\end{align} Combinig the latter equation with the first ODE you get the ODE system
\begin{align}
&\mathbf{f}( t, \mathbf{x}, \mathbf{z}) \, - \, \mathbf{x}' = 0 \\
&\frac{\partial \mathbf{g}}{\partial t}( t, \mathbf{x}, \mathbf{z}) + \frac{\partial \mathbf{g}}{\partial \mathbf{x}}( t, \mathbf{x}, \mathbf{z})\cdot \mathbf{f}( t, \mathbf{x}, \mathbf{z})   + \frac{\partial \mathbf{g}}{\partial \mathbf{z}}( t, \mathbf{x}, \mathbf{z}) \cdot\mathbf{z}' = 0
\end{align} 
If by any chance, the $m\times m$ matrix of derivatives $\frac{\partial \mathbf{g}}{\partial \mathbf{z}}( t, \mathbf{x}, \mathbf{z})$ is invertable for all values of the arguments $( t, \mathbf{x}, \mathbf{z})$ then the system turns into 
\begin{align}
\mathbf{x}'  & = \mathbf{f}( t, \mathbf{x}, \mathbf{z}) \\
\mathbf{z}' &= - \left(\frac{\partial \mathbf{g}}{\partial \mathbf{z}}( t, \mathbf{x}, \mathbf{z}) \right)^{-1}\left[\frac{\partial \mathbf{g}}{\partial t}( t, \mathbf{x}, \mathbf{z}) + \frac{\partial \mathbf{g}}{\partial \mathbf{x}}( t, \mathbf{x}, \mathbf{z})\cdot \mathbf{f}( t, \mathbf{x}, \mathbf{z})  \right]\\
\end{align} 
