Systems of differential equations

Eigenvectors turn the coupled system into two uncoupled scalar equations along the eigendirections: each $e^{\lambda_i t}\mathbf{v}_i$ moves purely along the line spanned by $\mathbf{v}_i$, growing or shrinking at rate $\lambda_i$, so the full motion is a blend of two independent straight-line motions.

First check $e^{\lambda_i t}\mathbf{v}_i$ solves the system: its derivative is $\lambda_i e^{\lambda_i t}\mathbf{v}_i$, while $A(e^{\lambda_i t}\mathbf{v}_i)=e^{\lambda_i t}A\mathbf{v}_i=e^{\lambda_i t}\lambda_i\mathbf{v}_i$ using $A\mathbf{v}=\lambda\mathbf{v}$; the two sides match, so it is indeed a solution for $i=1,2$.

Linear independence of $\mathbf{v}_1,\mathbf{v}_2$: suppose $c_1\mathbf{v}_1+c_2\mathbf{v}_2=\mathbf{0}$. Applying $A$ gives $c_1\lambda_1\mathbf{v}_1+c_2\lambda_2\mathbf{v}_2=\mathbf{0}$. Multiplying the first relation by $\lambda_2$ and subtracting gives $c_1(\lambda_1-\lambda_2)\mathbf{v}_1=\mathbf{0}$; since $\lambda_1\neq\lambda_2$ and $\mathbf{v}_1\neq\mathbf{0}$, we get $c_1=0$, and then $c_2=0$ too. So $\mathbf{v}_1,\mathbf{v}_2$ are independent.

By linearity of the system (exactly as in the scalar case), any combination $c_1e^{\lambda_1t}\mathbf{v}_1+c_2e^{\lambda_2t}\mathbf{v}_2$ solves the system. Because $\mathbf{v}_1,\mathbf{v}_2$ are independent, matching any initial condition $\mathbf{x}(0)=\mathbf{x}_0$ means solving $c_1\mathbf{v}_1+c_2\mathbf{v}_2=\mathbf{x}_0$ for $(c_1,c_2)$, which has a unique solution since $\{\mathbf{v}_1,\mathbf{v}_2\}$ is a basis of the plane. So every solution is captured, giving the general solution $\mathbf{x}(t)=c_1e^{\lambda_1t}\mathbf{v}_1+c_2e^{\lambda_2t}\mathbf{v}_2$.

The trace and determinant of $A$ are, respectively, the sum and product of its eigenvalues — so this theorem translates a statement about roots of a quadratic (both have negative real part) into a statement about two easily computed numbers, without ever solving for the eigenvalues explicitly.

By definition of the characteristic polynomial, $\lambda^2-(\text{tr}\,A)\lambda+\det A=0$ has roots $\lambda_{1,2}=\dfrac{\text{tr}\,A\pm\sqrt{(\text{tr}\,A)^2-4\det A}}{2}$, and by Vieta's formulas $\lambda_1+\lambda_2=\text{tr}\,A$ and $\lambda_1\lambda_2=\det A$.

If $\det A<0$, then $\lambda_1\lambda_2<0$, so the eigenvalues are real with opposite signs (a saddle); solutions grow without bound along the positive-eigenvalue direction, so the origin is unstable.

If $\det A>0$ and $\text{tr}\,A<0$: when the eigenvalues are real, $\lambda_1\lambda_2>0$ means they share a sign, and $\lambda_1+\lambda_2<0$ forces that sign to be negative, so both $e^{\lambda_1t}\to 0$ and $e^{\lambda_2t}\to 0$; when the eigenvalues are complex conjugates $\alpha\pm i\beta$, $\text{tr}\,A=2\alpha<0$ gives $\alpha<0$, so the amplitude factor $e^{\alpha t}\to 0$ while $\cos(\beta t),\sin(\beta t)$ stay bounded. In every sub-case, every solution decays to $\mathbf{0}$: asymptotic stability.

Conversely if $\text{tr}\,A>0$ (with $\det A>0$, so real eigenvalues share the trace's sign, or complex with positive real part), at least one exponential factor grows, so solutions starting near but not at the origin move away: instability. This covers every sign combination, proving the equivalence.