Eigenspaces of 2x2 Jacobian matrix with change of variable Let
$$
A=
\begin{pmatrix}
v & \rho\\
v\left(v+2\rho\right) & \rho\left(2v+\rho\right)
\end{pmatrix}
$$
where $\rho>0$, $v=\alpha-\rho$ with $\alpha\in\mathbb{R}$.
Show that $A$ has two distinct eigenvalues $\lambda_1(\rho,v)<\lambda_2(\rho,v)$ and compute a basis of eigenvectors assoiated to these eigenvalues.
My work
We have
$$\det(\lambda I_2-A)=\lambda^2-(v+\rho(2v+\rho))\lambda+v\rho(v-\rho)$$
The discriminant is
$$\Delta=v^2+4\rho^2v^2+\rho^4+4\rho^3v+6v\rho^2$$
I cannot seem to find a way to show that it is strictly positive. Even if I assume it is, I do not know how to find easily the eigenvectors (apart from resolving a system which will take a very long time). There has to be something that I am missing.
PS: this comes from an hyperbolic PDE where I need to show the system hyperbolicity:
$$
\left\{
\begin{array}{l}
\partial_t \rho+\partial_x(\rho(\alpha-\rho))=0\\
\partial_t(\rho\alpha)+\partial_x(\rho\alpha(\alpha-\rho))=0
\end{array}
\right.
$$
and we chose the change of variables ${\bf u} =(\rho,\alpha\rho)\to {\bf v} = (\rho,v)$ with $v=\alpha-\rho$.
 A: Note that we can write
$$
\Delta=v^2+4\rho^2v^2+\rho^4+4\rho^3v+6v\rho^2\\
= (1+4\rho^2)v^2+(4\rho^3+6\rho^2)v + \rho^4
$$
Consider the polynomial $p(v) = \Delta$. We see that this is a quadratic function with positive coefficients, so it will hold that $p(v) \geq 0$ for all $v$ if and only if its discriminant is either zero or negative. We find the discriminant of $p$ to be
$$
\Delta_v = (4\rho^3 + 6\rho^2)^2 - 4\rho^4(1 + 4 \rho^2) = 48\rho^5 + 32\rho^4.
$$
For positive $\rho$, $\Delta_v$ is positive. So, there must exist a value of $v$ for which $\Delta$ is negative, which in turn means that there are complex eigenvalues, contrary to your supposition.
A: The initial system reads \begin{aligned}
\rho_t + (\rho (\alpha-\rho))_x &= 0 \\
(\rho\alpha)_t + (\rho\alpha (\alpha - \rho))_x &= 0
\end{aligned}
for the conserved variables ${\bf u} = (\rho, \rho\alpha)$.
Now we introduce the new vector of conserved variables $
{\bf v} = (\rho, \alpha-\rho)
$. With $v=\alpha-\rho$, the first equation of our system rewrites as
$$
\rho_t + (\rho v)_x = 0
$$
and the second one becomes $$
(\rho (\rho+v))_t + (\rho v (\rho+v))_x = 0 \, .
$$
Using the first equation and the product rule, we may rewrite the latter as
$$
v_t + (v-\rho) v_x = 0 \, ,
$$
where we have also used $\rho\neq 0$.
Therefore, the Jacobian matrix of the flux such that ${\bf v}_t + A {\bf v}_x = \bf 0 $ reads $$
A = \begin{bmatrix}
v & \rho\\
0 & v-\rho
\end{bmatrix}
$$
which eigenvalues are the diagonal entries. They are distinct provided $\rho>0$, and the system is then strictly hyperbolic.
