Why is a linear autonomous system asymptotically stable iff for all eigenvalues $\lambda$ of $A$, $Re(\lambda) < 0$

Question

I'm trying to understand asymptotic stability of linear antonymous systems. I'm not sure if for the system $x' = Ax$, $x(t) = 0$ is the only fixed point that can be stable.

In any case, I can imagine that any state, x, can be represented as a sum of eigenvectors, $v_1 + v_2 + ... + v_n$, of $A$ plus some other vector, $u$, that is not an eigenvector. If all eigenvalues, $\lambda$, of $A$ satisfy $Re(\lambda) < 0$, then the eigenvector summands of $x$ will aproach $0$ as $t\rightarrow\infty$. However, that leaves the behavior of $u$ unexplained. Perhaps it can be shown that over time $u$ must evolve into an eigenvector of $A$? Or maybe I'm totally off the mark here.

Answer 1

Use Jordan canonical form. For a $k \times k$ Jordan block $B = \lambda I + N$, where $N^k = 0$, $\exp(tB) = \exp(t\lambda) \exp(tN)$ where $\exp(tN) = I + tN + \ldots + t^{k-1} N^{k-1}/(k-1)!$ is polynomial in $t$. Thus if $\text{Re}\; \lambda < 0$, $\exp(tB) \to 0$ as $t \to \infty$.

Answer 2

Well, $u$ would be a generalized eigenvector in your case. Then the solution of $\dot{x} = Ax$ would be $$ x(t) = \sum_{i=1}^{k} \sum_{j=1}^{m_i} c_{ij} v_{ij} \frac{t^{\ell_{ij}-1}}{(\ell_{ij}-1)!} e^{\lambda_i t}$$

where $k$ is the number of different eigenvalues of $A$, $m_i$ is the algebraic multiplicity of the eigenvalue $\lambda_i$, $v_{ij}$ is the (generalized) eigenvectors of $A$ such that

$$ (A - \lambda_i I) ^ {\ell_{ij}} v_{ij} = 0 $$

but

$$ (A - \lambda_i I) ^ {\ell_{ij}-1} v_{ij} \neq 0 $$

and $c_{ij}$ depend on the initial condition. Note that $\sum_{i=1}^{k} m_i = n$ and $\ell_{ij} \leq m_i$ where $n$ is the system order.

This general case may seem complicated but it is easier when working on a specific example. The general idea is that the effect of the "other vectors" comes in polynomially in time but the effect of $\lambda_i$ comes in exponentially, therefore the result in the question.