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.
 A: I suppose that the given system has real coefficients.
That is the trick I remember:
$\dot x=a_{11}x+a_{12}y\implies\ddot x=a_{11}\dot x+a_{12}\dot y\implies\ddot x=a_{11}\dot x+a_{12}(a_{21}x+a_{22}y)\implies\\\ddot x=a_{11}\dot x+a_{12}a_{21}x+a_{22}a_{12}y\implies\ddot x=a_{11}\dot x+a_{12}a_{21}x+a_{22}(\dot x-a_{11}x)\implies\\\ddot x=(a_{11}+a_{22})\dot x+(a_{12}a_{21}-a_{11}a_{22})x$
Hence we get a $2$nd order linear differential equation
$$\ddot x-(a_{11}+a_{22})\dot x+(a_{11}a_{22}-a_{12}a_{21})x=0\tag{LDE}$$
with characteristic equation $m^2-Tr(A)m+\Delta(A)=0$.
We have two cases:

*

*There are two real roots, $\lambda_1,\lambda_2$ and the solution of the LDE is $x(t)=c_1e^{\lambda_1t}+c_2e^{\lambda_2t}$. For asympotic stability (AS), $x(t)\rightarrow 0$ when $t\rightarrow\infty$. Assuming the general case of the solution, this happens iff $\lambda_1,\lambda_2<0$ iff $Tr(A)=\lambda_1+\lambda_2<0$ and $\Delta(A)=\lambda_1\lambda_2>0$.

*There are two complex-conjugate roots $\alpha\pm\beta i$ and the solution is $x(t)=e^{\alpha}(c_1\cos\beta t+c_2\sin\beta t)$. Again, the solution is AS iff $\alpha=Tr(A)/2<0\implies Tr(A)<0.$ In this case, $\Delta(A)=\alpha^2+\beta^2>0.$
A: 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
$$ a_{11} a_{22} - a_{12} a_{21} > 0. $$
