Stability of linear systems with complex eigenvalues

Question

I'm very confused with stability of linear systems, especially when they have complex eigenvalues. Suppose I have the matrix

$$\begin{bmatrix}-5 & 3\\-1&1\end{bmatrix}$$

whose eigenvalues are $-2+i\sqrt{6}$ and $-2-i\sqrt{6}$. This is an unstable system, but can someone explain why? My book has a theorem but I don't understand what it's saying.

Also, if I have a system with one negative eigenvalue and one that is zero, does that make it stable? I know negative eigenvalues are stable but not sure about the zero. Thanks!

Answer 1

• If any eigenvalue has a positive real part, the system will tend to move away from the fixed point (unstable system).
• If any eigenvalue has a negative real part, the system will tend to move back to steady state (stable system).
• If any eigenvalue has an imaginary part, the system oscillate around the steady state.
• If eigenvalue is zero, the system remains position or amplitude constant.

This is because find eigenvalues is similar to find roots of auxiliar equation, where roots are arguments of exponential function times independent value.

$$ f(t) = \sum_{i=0}^n A_{i}\exp(\lambda_{i}t). $$

Answer 2

It is because you have to find the modulus of the complex Eigen value not look at the positive and negative individually

if your Eigen value is 1+- 3i the modulus is sqrt(10)

sqrt(10)>1 thus the system is unstable

your intuition tells you the 1-3i would be < 1 and be stable but that is not the case

in higher order math the complex root tells you more about the behavior, but in you basic is Eigen value greater than equal to or less than 1 hence unstable you look at the modulus of a complex root

its the Eigen values "length" in the complex world

Answer 3

The real parts of your complex eigenvalues are negative, therefore the system is stable not unstable as you suggest.