Controllability of a convex polytope of matrices of LTI system

Given Linear time invariant (LTI) dynamic system:

\begin{align*} \dot{x}(t)=Ax(t)+Bu(t) \end{align*} where $A \in R^{n \times n}$ and $B \in R^{n}$ are system matrices, $x(t)$ is the system's state and $u(t)$ is the system's control input.

The question is to show if the following is controllable or not:

If $A=A_0+\sum_{i=1}^{p} w_i A_i$, where $A_0$ is Hurwitz (stable matrix), $A_i=B \theta_i^T \in R^{n \times n}$, $\theta_i \in R^{n}$, and $\sum_{i=1}^{p} w_i=1$, $w_i \ge 0$, what can be said about the controllability of the pair $(A,B)$, given that $(A_i,B)$ is controllable for $\forall i={0,\cdots, p}$?

Any ideas how to proceed? Thanks.

• Given your definition for $A_i$ for $i>0$ it is impossible for the pairs $(A_i,B)$ to be controllable unless $n=1$. So since it is given that all $(A_i,B)$ are controllable implies that $n=1$, so any nonzero $B$ should suffice. – Kwin van der Veen Jun 26 '18 at 1:38
• @KwinvanderVeen Can you elaborate plz? – ems Jun 26 '18 at 1:47
• $A_i^k\,B = B\,(\theta_i^\top\,B)^k$ so every column of the controllability matrix lies in the span of $B$. – Kwin van der Veen Jun 26 '18 at 2:07
• you mean, $A_i^kB = (B \theta_i^T)^kB$? – ems Jun 26 '18 at 2:10
• Are those different? – Kwin van der Veen Jun 26 '18 at 2:11

Given your definition for $A_i$ for $i>0$ it is impossible for the pairs $(A_i,B)$ to be controllable, unless $n=1$. Namely for $A_i^2\,B$ the following factorization is true
$$B\,\theta_i^\top\,B\,\theta_i^\top\,B = B\,(\theta_i^\top\,B)\,(\theta_i^\top\,B) = B\,(\theta_i^\top\,B)^2$$
and for higher powers it can also be shown that $A_i^k\,B=B\,(\theta_i^\top\,B)^k$, therefore each column of the controllability matrix $(A_i,B)$ will lie in the span of $B$. So the rank will be equal to the rank of $B$. So since it is given that all $(A_i,B)$ are controllable implies that $n=1$, so any nonzero $B$ should suffice.
If any $A_i$ would be allowed then in general nothing can be said about the controllability of $(A,B)$. For example when $p=1$ and $(A_0,B)$ with $A_0$ Hurwitz, then by using $A_1 = -A_0$, since $(\alpha\,A_0,B)$ for nonzero $\alpha$ is also controllable, one gets $A=0$ which means $(A,B)$ is uncontrollable if $n>1$. When using any other nonzero value for $\alpha$ then $(A,B)$ will be controllable.