Putting an ansatz in a system, and getting an equation for $\lambda$ Consider
$$
\frac{\partial u}{\partial t}=fv-g\frac{\partial\eta}{\partial x}\\
\frac{\partial v}{\partial t}=-fu-g\frac{\partial\eta}{\partial y}\\
\frac{\partial\eta}{\partial t}=-H_0\frac{\partial u}{\partial x}-H_0\frac{\partial v}{\partial y}.
$$
Here $u,v$ and $\eta$ are functions of $(x,y,t)$, $f\in\mathbb{R}$ is a parameter, $g$ is the gravitational acceleration and $H_0$ is a constant.
Now make the ansatz
$$
(u,v,\eta)^T=e^{i(kx+ly)+\lambda t}\cdot (u_0,v_0,\eta_0)^T.
$$


Now it is said that inserting this ansatz into the equation, we get 
    $$
0=\lambda(\lambda^2+gH_0(k^2+l^2)+f^2).
$$


How to get this?
If I insert the ansatz in the system, I only get three equations for $\lambda$, namely
$$
\lambda=u_0^{-1}(fv_0-ikg),\\
\lambda=v_0^{-1}(-fu_0-ilg\eta_0),\\
\lambda=\eta_0^{-1}(H_0u_0ik-H_0v_0il)
$$
--- By the way: Where does tis ansatz come from?
Is it a product ansatz 
$$
(u(x,y,t),v(x,y,t),\eta(x,y,t))=(u(t),v(t),\eta(t))\cdot (u(x,y),v(x,y),\eta(x,y))
$$
and choosing the space-part $(u(x,y),v(x,y),\eta(x,y))$ to be eigenmodes/ Fouriermodes oscillating with same spatialbehaviour? I.e. is the idea maybe to write solutions in fouriermode basis and tehrefore to start with fouriermodes as solutions (for the space-dependent part)?
 A: We have a system of three equations:
\begin{equation}
\begin{cases}
u_0\lambda=fv_0-g\eta_0ik\\v_0\lambda=-fu_0-g\eta_0il\\ \eta_0\lambda=-H_0u_0ik-H_0v_0il
\end{cases}
\begin{cases}
u_0={fv_0-g\eta_0ik\over\lambda}\\v_0\lambda=-fu_0-g\eta_0il\\ \eta_0\lambda=-H_0u_0ik-H_0v_0il
\end{cases}
\end{equation}
we get $u_0$ from the first equation and we replace it in the other ones, so:
\begin{equation}
\begin{cases}
u_0={fv_0-g\eta_0ik\over\lambda}\\v_0\lambda={-f^2v_0+fg\eta_0ik\over\lambda}-g\eta_0il\\ \eta_0\lambda=-H_0{fv_0-g\eta_0ik\over\lambda}ik-H_0v_0il
\end{cases}
\end{equation}
Now we make some manipulation with the second and third equation:
\begin{equation}
\begin{cases}
v_0(\lambda^2+f^2)=\eta_0(fgik-\lambda gil)\\ \eta_0(\lambda^2+H_0k^2g)=v_0(-H_0ikf-\lambda H_0il)
\end{cases}
\begin{cases}
{v_0\over\eta_0}={(fgik-\lambda gil)\over(\lambda^2+f^2)}\\ {\eta_0\over v_0}={(-H_0ikf-\lambda H_0il)\over(\lambda^2+H_0k^2g)}
\end{cases}
\begin{cases}
{v_0\over\eta_0}={gi(fk-\lambda l)\over(\lambda^2+f^2)}\\ {\eta_0\over v_0}={-H_0i(kf+\lambda l)\over(\lambda^2+H_0k^2g)}
\end{cases}
\end{equation}
Now we multiply these equations:
\begin{equation}
{v_0\over\eta_0}{\eta_0\over v_0}={gi(fk-\lambda l)\over(\lambda^2+f^2)}{-H_0i(kf+\lambda l)\over(\lambda^2+H_0k^2g)}\longrightarrow 1=gH_0{(f^2k^2-\lambda^2l^2)\over(\lambda^2+f^2)(\lambda^2+H_0k^2g)}
\end{equation}
Hence we get:
$$\lambda^4+\lambda^2H_0k^2g+f^2\lambda^2+f^2H_0k^2g=gH_0f^2k^2-gH_0l^2\lambda^2$$
$$\lambda^4+\lambda^2H_0k^2g+f^2\lambda^2+gH_0l^2\lambda^2=0$$
$$\lambda^2[\lambda^2+gH_0(k^2+l^2)+f^2]=0$$
