# Linear stability analysis on a constrained three-dimensional system of ODE

Let $\begin{cases} \dot x = f({\bf u}) \\ \dot y = g({\bf u}) \\ \dot z = h({\bf u})\end{cases}$ be a well-defined nonlinear system with ${\bf u} = (x,y,z)$ and restricted to domain $x,y,z \geq 0$. Suppose that $f+g+h=0$ so that $\dot x + \dot y + \dot z = 0 \Rightarrow x + y + z = K$ constant. I want to perform a linear stability analysis of some fixed point $\bf u_0$ but when I calculate the Jacobian and find eigenvalues I get a $0$ eigenvalue. I would like to know:

1. If I calculate the Jacobian of the system in $x,y$ and I get eigenvalues with nonzero real part, can I use the result to classify the stability of $\bf u_0$?
2. If I calculate the Jacobian in $x,y$, $y,z$, and $x,z$ and get the same result (eigenvalues with nonzero real part) can I use the result to classify the stability of $\bf u_0$?
3. Is it necessarily the case that a constrained system as described would have a $0$ eigenvalue in the Jacobian?

The zero eigenvalue is directly related to the fact that $\dot x+\dot y+\dot z=0$. If you make the substitution $z=K-x-y$, then you can look at the jacobian in terms of $x$ and $y$ directly. That is,
$$\dot x = f(x,y,K-x-y)\\ \dot y = g(x,y,K-x-y)$$ Now, the Jacobian is
$$J=\left(\begin{matrix}f_x-f_z&f_y-f_z\\g_x-g_z&g_y-g_z\end{matrix}\right)$$ and the eigenvalues satisfy the characteristic polynomial $$\lambda^2-(f_x-f_z+g_y-g_z)\lambda+(f_x-f_z)(g_y-g_z)-(f_y-f_z)(g_x-g_z)$$ Notice that the coefficient of $\lambda$ is equal to $-(f_x+g_y+h_z)$. Indeed, if you look at the characteristic polynomial for the original system, you'll find that it's equal to this characteristic polynomial multiplied by $\lambda$. Therefore, you will get the same eigenvalues other than $\lambda=0$.
Thus, you can say that the stability is governed by the eigenvalues other than $\lambda=0$.