# Are uncontrollable modes with no real part stabilizable?

so this is a simple question regarding stabilizability.

When I want to know if a certain uncontrollable system is stabilizable, I need to find the uncontrollable modes (also known as poles, or eigenvalues) and see if they are in the left-half complex plane.

But I want to know about those modes that don't have a real part, so they lie over the axis, these are usually named marginally stable modes. How do this modes interact with stabilizability?

Thanks!

• The essence of stabilizability is that all uncontrollable modes are stable and all unstable modes are controllable. If $${\text{rank}}\left[ {\begin{array}{*{20}{c}} {\lambda I - A}&B \end{array}} \right] = n$$ for all $\lambda$ with Re$(\lambda) \geq 0$ then system is stabilizable. That includes marginally stable poles. – ITA Oct 20 '16 at 16:49

## 1 Answer

A state space model is called stabilizable if there exists some $u(t)$ such that,

$$\lim_{t\to 0}x(t)=0\quad\forall\ x(0)\in \mathbb{R}^n.$$

So if $(A,B)$ has marginally stable modes, which are unobservable, then those modes can never die out and therefore $x(t)$ will not go to zero for all initial conditions. Such a system would therefore not be stabilizable.

In order words, using the Hautus test, $(A,B)$ is stabilizable if and only if,

$$\text{rank}\left[\lambda\,I-A \quad B\right]=n \quad \forall \ \ \text{Re}(\lambda) \geq 0,$$

where you only need to fill in eigen values of $A$ for $\lambda$, since otherwise $\lambda\,I-A$ is already full rank.