Stability of linear systems via eigenvalues

Statement

Let $A$ be a real $n \times n$ matrix and consider the linear system with constant coefficients $x' = A x$ on $\mathbb{R}^n$, with the equilibrium point $0$. (1) The point $0$ is asymptotically stable if and only if each eigenvalue $\lambda$ of $A$ has $\operatorname{Re} \lambda < 0$. (2) If some eigenvalue $\lambda$ of $A$ has $\operatorname{Re} \lambda > 0$, then $0$ is unstable. (3) The point $0$ is Lyapunov stable if and only if two conditions hold. Each eigenvalue has $\operatorname{Re} \lambda \le 0$. Each eigenvalue with $\operatorname{Re} \lambda = 0$ has equal algebraic and geometric multiplicities.

Remarks

The proof uses the matrix exponential. The entries of $e^{tA}$ are sums of terms $t^k e^{\lambda t}$. Here $\lambda$ is an eigenvalue and $k$ is less than the size of its largest Jordan block. In case (1), there are constants $C, \alpha > 0$ with $\|e^{tA}\| \le C e^{-\alpha t}$ for $t \ge 0$. So the origin is globally and exponentially stable. For $A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$, both eigenvalues are $0$ but the solution $(t, 1)$ is unbounded. So $0$ is unstable, in accordance with (3). The Routh–Hurwitz criterion tests the condition of (1) from the coefficients of the characteristic polynomial, without computing the roots.

Explanations

L1 human

For a linear system, the eigenvalues decide everything. Eigenvalues with negative real part give decaying solutions, and an eigenvalue with positive real part gives a growing solution.