The critical point $(0,0)$ is assymptotically stable. Demostrate $a_{11}+a_{22}<0$ and $a_{11}a_{22}-a_{12}a_{21}>0.$ I've got the following linear system:
$$\frac{dx}{dt}=a_{11}x+a_{12}y$$
$$\frac{dy}{dt}=a_{21}x+a_{22}y$$
The critical point $(0,0)$ is an assymptotically stable critical point of the system.
We have to demostrate that $$a_{11}+a_{22}<0$$ and $$a_{11}a_{22}-a_{12}a_{21}>0.$$

I have writted the system like this:
$$\begin{pmatrix}
x' \\
y'
\end{pmatrix}=\begin{pmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22} \\
\end{pmatrix}\begin{pmatrix}
x \\
y
\end{pmatrix}$$
If we call $A=\begin{pmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22} \\
\end{pmatrix}$
$Traza(A)=a_{11}+a_{22}$
and
$Det(A)=a_{11}a_{22}-a_{12}a_{21}$
So now, I have to do something with the eigenvalues, true?
How can I follow? I'm a bit lost.

Let's take an alternative approach without solving for the eigenvalues. The phrase "asymptotically stable" implies that Lyapunov's method can be used, in which the following conditions must be satisfied:

*

*$ V\left(\mathbf{0}\right) = 0 $

*$ V\left(\mathbf{x}\right) > 0, \quad \forall \; \mathbf{x} \neq \mathbf{0} $

*$ \dot{V}\left(\mathbf{x}\right) < 0, \quad \forall \; \mathbf{x} \neq \mathbf{0} $
Consider the Lyapunov function candidate:
$$ V\left(\mathbf{x}\right) = - \frac{1}{2} a_{21} x^{2} + \frac{1}{2} a_{12} y^{2}. $$
The first condition is satisfied when $(x, y) = (0, 0)$. In order to satisfy the second condition $V\left(\mathbf{x}\right) > 0$, then we must have
$$ a_{12} > 0, $$
$$ a_{21} < 0. $$
Taking the time derivative for $V\left(\mathbf{x}\right)$, we have
$$ \dot{V}\left(\mathbf{x}\right) = - a_{21} x \dot{x} + a_{12} y \dot{y} $$
$$ \dot{V}\left(\mathbf{x}\right) = - a_{21} x \left(a_{11} x + a_{12} y\right) + a_{12} y \left(a_{21} x + a_{22} y\right) $$
$$ \dot{V}\left(\mathbf{x}\right) = - a_{11} a_{21} x^{2} - a_{12} a_{21} x y + a_{12} a_{21} x y + a_{12} a_{22} y^{2}. $$
Cancelling like terms yields
$$ \dot{V}\left(\mathbf{x}\right) = - a_{11} a_{21} x^{2} + a_{12} a_{22} y^{2}. $$
In order to satisfy the third condition $\dot{V}\left(\mathbf{x}\right) < 0$, then we must have
$$ a_{11} a_{21} > 0, $$
$$ a_{12} a_{22} < 0. $$
From $a_{12} > 0$ and $a_{21} < 0$, it must be true that
$$ a_{11} < 0, $$
$$ a_{22} < 0. $$
Since $a_{11} < 0$ and $a_{22} < 0$, then
$$ a_{11} + a_{22} < 0. $$
Similarly, since $a_{11} a_{22} > 0$ and $a_{12} a_{21} < 0$, then
$$ a_{11} a_{22} - a_{12} a_{21} > 0. $$
