Prove that all of the solutions of a 2 x 2 dynamical system approaches the origin iff these three conditions are true Prove that if  $X_{n+1} = AX_n$, where $X_n \in \Bbb R^2$ and $A$ is a 2 x 2 matrix all the solutions  of the planar system  approaches the origin when $n \to \infty$ iff:
$\DeclareMathOperator{\Tr}{Tr}$
$1-\Tr A + \det A > 0$
$1+ \Tr A + \det A > 0$
$1 - \det A > 0$
My try:
I'm studying dynamical systems of 2x2 where I only want to study the stability of the origin as the only fixed point.
I was taught that, in dynamical systems of 2x2, the stability of the origin is given by 
$\rho(A)$=$\max\{|\lambda_1| , |\lambda_2| \}$, where $\lambda_{1,2}$ are the eigenvalues of the matrix $A$
If $\rho(A) < 1$, the origin is stable, hence, attractor.
If $\rho(A) > 1$, the origin is unstable.
I tried to apply this criteria to study those three conditions, using the characteristic polynomial associated to the matrix $A$, solving the quadratic and analyzing by cases ($\lambda_1 \ne \lambda_2$ and $\lambda_{1,2} > 0$), etc but I don't know how to proceed.
Any hints? 
 A: As a first note, you have the conditions on $\rho(A)$ copied backwards---it should read that if $\rho(A) <1$ then the origin is stable (attractor) and similarly if $\rho(A) > 1$ then the origin is unstable. Consider the action of $A$ on a corresponding eigenvector.
As for your main question, consider an arbitrary $2 \times 2$ matrix:
$$A := \begin{bmatrix} a & b \\ c & d \end{bmatrix}.$$
We wish to find the eigenvalues of $A$ via the characteristic polynomial:
\begin{align*}
\det(A - \lambda I) &= 0 \\
\begin{vmatrix} a - \lambda & b \\ c & d-\lambda \end{vmatrix} &= 0 \\
(a-\lambda)(d-\lambda) - bc &= 0 \\
\lambda^2 -(a+d) \lambda + ad - bc &=0 \\
\lambda^2 - \mathrm{tr}(A) \lambda + \det(A) &=0.
\end{align*}
This is actually a surprisingly handy formula to memorize---we can always get to the characteristic polynomial with just the trace and determinant; no need to keep recomputing it each time. Continuing onwards with the quadratic equation, we find
$$ \lambda = \frac{\mathrm{tr}(A) \pm \sqrt{ (\mathrm{tr}(A) )^2 - 4 \det(A) }}{2} .$$
As we want the origin to be stable, we need $\rho(A)<1$, i.e.,
$$1 > \left\vert \frac{ \mathrm{tr}(A) \pm \sqrt{ (\mathrm{tr}(A) )^2 - 4 \det(A) } } {2}\right\vert$$
or
$$2 > \left\vert \mathrm{tr}(A) \pm \sqrt{ (\mathrm{tr}(A) )^2 - 4 \det(A) } \right\vert.$$
Breaking apart the absolute value, this becomes
$$2 >  \mathrm{tr}(A) \pm \sqrt{ (\mathrm{tr}(A) )^2 - 4 \det(A) } > -2.$$
Remembering that $\sqrt{ (\mathrm{tr}(A) )^2 - 4 \det(A) } \geq 0$, we can rewrite this as two inequalities:
$$2>  \mathrm{tr}(A) + \sqrt{ (\mathrm{tr}(A) )^2 - 4 \det(A) } \quad \text{and}\quad  \mathrm{tr}(A) - \sqrt{ (\mathrm{tr}(A) )^2 - 4 \det(A) } > -2.$$
Rearranging these gives you the first two inequalities.
As for the final inequality, recall that the determinant is equal to the product of the eigenvalues:
$$\det(A) = \lambda_1 * \lambda_2.$$
If $|\lambda_1| < 1$ and $|\lambda_2 |<1$, then
$$ -1 < \lambda_1 * \lambda_2 = \det(A) < 1$$
which can easily be rearranged to give the third inequality. Of note is that adding the first two inequalities together yields that $2 + 2 \det(A) >0$, so there is no need to worry about both sides for this final inequality.
Edit: Complex Eigenvalues
In response to the comment, I wish to address the possibility of $\lambda_1, \lambda_2 \in \mathbb{C}$.
Recall that for a $2 \times 2$ matrix, complex eigenvalues will be conjugate, that is,
$$\lambda_1 = \alpha + i \beta \quad \lambda_2 = \alpha - i \beta$$
for some $\alpha, \beta \in \mathbb{R}$. Then we have that the spectrum $\rho(A) < 1$ if and only if $\alpha^2 + \beta^2 < 1$. However, we have explicit formulas for $\alpha$ and $\beta$ from finding the roots of the characteristic polynomial:
\begin{align*}
\alpha &= - \frac{\mathrm{tr}(A)}{2} \\
\beta &= \frac{\sqrt{4 \det(A) - (\mathrm{tr}(A))^2}}{2}, 
\end{align*}
noting that we have factored out $\sqrt{-1} = i$ in the discriminant for the $\beta$ term. Thus
\begin{align*}
1 &> \alpha^2 + \beta^2 \\
 &= \frac{ (\mathrm{tr}(A))^2}{4} + \frac{ 4 \det(A) - (\mathrm{tr}(A))^2}{4}
\end{align*}
which simplifies to
$$1 - \det(A) > 0,$$
which is precisely the third inequality.
