Equilbrium points in dynamical systems I have been giving the following system of equations. From this system of equations I have determined the jacobian matrix, which i then worked out. However I am wondering what is the equilibrium points of this system? I assume $(0,0,0,0,0,0,0)$ is one. So my question are how do I evaluate this into my workings below and find out its stability? And is there another equilibrium point?
\begin{eqnarray}
\frac{dV_H}{dt} &=&- \left( \mu_H + \mu_{AVD} + \mu_{AND} + w \right)V_H  + \gamma e S_H \label{equation1}\\
\frac{dS_H}{dt}  &=& -bp_HI_M\frac{S_H}{N_H} + wV_H - \left( \gamma e  + \mu_H  \right)S_H  \label{equation2}\\
\frac{dI_H}{dt} &=&  - \left( \mu_H + \mu_{YF} + r \right)I_H + bp_H I_M \frac{S_H}{N_H}\label{equation3}\\
\frac{dR_H}{dt} &=& -\mu_HR_H  + rI_H \label{equation4}\\
\frac{dS_M}{dt} &=& -bp_MS_M\frac{I_H}{N_H} - \mu_M S_M + \alpha_M I_M + \alpha_M E_M \label{equation5}\\
\frac{dE_M}{dt} &=& - \left( \beta + \mu_M  \right) E_M + bp_M S_M  \frac{I_H}{N_H} \label{equation6}\\
\frac{dI_M}{dt} &=& -\mu_MI_M+\beta E_M \label{equation7}
\end{eqnarray}
From this system i have determine the jacobian matrix to be the following
\begin{bmatrix}
\frac{\partial V_H}{\partial V_H}    & \frac{\partial V_H}{\partial S_H} &  \frac{\partial V_H}{\partial I_H} &  \frac{\partial V_H}{\partial R_H} &  \frac{\partial V_H}{\partial S_M} &  \frac{\partial V_H}{\partial E_M} &  \frac{\partial V_H}{\partial I_M} \\ 
  \frac{\partial S_H}{\partial V_H}    & \frac{\partial S_H}{\partial S_H} &  \frac{\partial S_H}{\partial I_H} &  \frac{\partial S_H}{\partial R_H} &  \frac{\partial S_H}{\partial S_M} &  \frac{\partial S_H}{\partial E_M} &  \frac{\partial S_H}{\partial I_M} \\ 
   \frac{\partial I_H}{\partial V_H}    & \frac{\partial I_H}{\partial S_H} &  \frac{\partial I_H}{\partial I_H} &  \frac{\partial I_H}{\partial I_H} &  \frac{\partial I_H}{\partial S_M} &  \frac{\partial I_H}{\partial E_M} &  \frac{\partial I_H}{\partial I_M} \\ 
    \frac{\partial R_H}{\partial V_H}    & \frac{\partial R_H}{\partial S_H} &  \frac{\partial R_H}{\partial I_H} &  \frac{\partial R_H}{\partial R_H} &  \frac{\partial R_H}{\partial S_M} &  \frac{\partial R_H}{\partial E_M} &  \frac{\partial R_H}{\partial I_M} \\ 
        \frac{\partial S_M}{\partial V_H}    & \frac{\partial S_M}{\partial S_H} &  \frac{\partial S_M}{\partial I_H} &  \frac{\partial S_M}{\partial R_H} &  \frac{\partial S_M}{\partial S_M} &  \frac{\partial S_M}{\partial E_M} &  \frac{\partial S_M}{\partial I_M} \\  
        \frac{\partial E_M}{\partial V_H}    & \frac{\partial E_M}{\partial S_H} &  \frac{\partial E_M}{\partial I_H} &  \frac{\partial E_M}{\partial R_H} &  \frac{\partial E_M}{\partial S_M} &  \frac{\partial E_M}{\partial E_M} &  \frac{\partial E_M}{\partial I_M} \\  
    \frac{\partial I_M}{\partial V_H}    & \frac{\partial I_M}{\partial S_H} &  \frac{\partial I_M}{\partial I_H} &  \frac{\partial I_M}{\partial R_H} &  \frac{\partial I_M}{\partial S_M} &  \frac{\partial I_M}{\partial E_M} &  \frac{\partial I_M}{\partial I_M} 
    \end{bmatrix}
Working out each partial derivative i have produced the following.
\begin{bmatrix}
    A    & \gamma e & 0 & 0 & 0 & 0 & 0 \\ 
    w      & B & 0 & 0 & 0 & 0 & -bp_H \frac{S_H}{N_H} \\ 
    0      & bp_H \frac{I_M}{N_H} & C & 0 & 0 & 0 & bp_H \frac{S_H}{N_H} \\ 
    0      & 0 & r  & D & 0 & 0 & 0 \\ 
        0      & 0 & -bp_H \frac{S_M}{N_H}  & 0 & E & \alpha_M & \alpha_M \\ 
        0      & 0 & bp_M \frac{S_M}{N_H}  & 0 & bp_M \frac{I_H}{N_H} & F & 0 \\ 
      0      & 0 & 0  & 0 & 0 & \beta & G
    \end{bmatrix}
where
\begin{align*}
A&=-\mu_H-\mu_{H-AVD}-\mu_{H-AND} - w \\
    B&=bp_H \frac{I_M}{N_H} - \gamma e -\mu_H\\
    C&= - \left( \mu_H + \mu_{H-YF} +  r \right)\\
    D&=-\mu _ H \\
    E&=-bp_M \frac{I_H}{N_H} - \mu_M\\
    F&=-\beta -\mu_M\\
    G&=-\mu _M
    \end{align*}
 A: So far all these steps are reasonable. 
To determine whether the zero equilibrium point is stable, you may have to find the spectrum of the linearization for special parameter values, or you could employ the Routh-Hurwitz criterion which will give you a set of algebraic equations for the coefficients.
There is indeed a second equilibrium point. All its components are proportional to $N_H$. For example,
$$
I_M = - \frac{(\gamma  e+\mu_H) (\mu_{AND}+\mu_{AVD}+\mu_H)+\mu_H
   w}{bp_H (\mu_{AND}+\mu_{AVD}+\mu_H+w)} \cdot N_H
$$
To find all other components, use a computer algebra system. 
Note that if all parameters are positive, then $I_M < 0$ for this equilibrium. 
Added: You can find the other components of the second equilibrium point as follows. Set all terms on the right hand side equal to 0. The last equation gives $E_M$. The second to last equation then tells you what $S_MI_H$ must be. The fifth equation then can be used to find $S_M$ and the value of $I_H$ follows. Next use the fourth equation to find the value of $R_H$. Then the third equation results in $S_H$, and the second equation finally gives you $V_H$. 
The first equation can be used as a consistency check.
Added 2: The linearization of this system about the zero equilibrium point is block diagonal. Using your notation, the top left $2 \times 2$ block is
$\begin{pmatrix} A & \gamma e \\ w & B 
\end{pmatrix}$. 
This matrix has the characteristic polynomial $z^2 - (A + B)z + (AB - \gamma e \cdot w)$. Both coefficients are positive, therefore both eigenvalues of this matrix are in the left half plane. That's the Routh Hurwitz criterion for $2 \times 2$ matrices.
The second block is $\begin{pmatrix} C & 0 \\ r & D 
\end{pmatrix}$ with characteristic polynomial $(z + C)(z+D)$, hence both eigenvalues of this matrix are also in the left half plane.
The third block is $3\times 3$, namely $\begin{pmatrix} E & \alpha_M & \alpha_M \\ 0  & F & 0 \\ 0 & \beta & G \end{pmatrix}$ and its characteristic polynomial is $(z+E)(z+F)(Z+G)$.  Again all its roots are in the left half plane.
The set of eigenvalues of the entire $7 \times 7$ matrix now is the union of the sets of eigenvalues of those three blocks, so they are all in the left half plane. That means that the zero equilibrium is always locally stable.
