# Why do real positive eigenvalues result in an unstable system? What about eigenvalues between 0 and 1? or 1?

I'm confused why, if all eigenvalues of a linear system are real and positive, this entails an unstable system. For example, if eigenvalues are between 0 and 1, surely this means the system is gradually shrinking? And even more so, doesn't an eigenvalue of 1 mean the system is "staying put"? Why is a negative eigenvalue instead related to stability?

I'm confused because in the case of eigenvalues of a Markov matrix, it seemed an eigenvalue of 1 meant stability, and 0 < λ < 1 meant it was shrinking. But in both those cases the eigenvalue is positive.

For a continuously differentiable dynamical system, a local perturbation around the evolution of the system near a fixed point is described by an exponential $$e^{\lambda t}$$ since (taking liberal approximations) $$x(t+\Delta t) \approx x(t) + \lambda x(t) = (1 + \lambda)x(t)$$; the function grows if $$\lambda > 0$$ and shrinks if the opposite.
For a discrete-time system like a Markov chain, local perturbations grow depending on whether or not the "next" step of the system is larger than the previous one, a.k.a. if $$x_{n+1} = \lambda x_{n} > x_n$$. Hence, $$\lambda > 1$$ indicates an unstable fixed point.
If the “system” you’re asking about is a system of differential equations, then instead of the behavior of successive powers of a matrix $$A$$ we’re interested in its exponential $$e^{tA}$$, which in turn depends on exponentials $$e^{\lambda t}$$of its eigenvalues. Here, it’s eigenvalues with a positive real part whose contributions grow without bound, while the contributions of ones with a negative real part tend to zero.