Let $A\in\mathbb{R}^{n\times n}$ be symmetric positive definite and $B\in\mathbb{R}^{n\times m}$. Construct the complex matrix $$Z=BB^{\prime} +jA \text{ where } j^2=-1.$$ Consider now the following system $$ \dot{x}= Ax+Bu.$$
I want to show that:
- The system is controllable if $Z$ has no eigenvalue on the imaginary axis.
- $Z$ has no eigenvalue on the imaginary axis if the system is controllable.
I tried to use the following theorem from "Linear Systems Theory, by João P. Hespanha", but I didn't get the result, yet. I think Sylvester equation that may help.
I will appreciate your help!
Suppose $Z$ has an eigenvalue on the imaginary axis $ja$ where $a$ is real, and its associated unit-length left-eigenvector is $w \in \mathbb{C}^n$. So, $$w'Z = w'BB' + jw'A = j a w'$$ Since $A > 0$ there exists $L$ such that $A = LL'$. If we multiply with $w$ from right we get $$w'BB'w + jw'LL'w = \lVert B'w \rVert + j \lVert L'w \rVert = j a$$ Therefore, $w'B=0$ and so $w'A = a w'$ from the first equation. So a left eigenvector of $A$ is orthogonal to the columns of $B$, which means the system is not controllable per the PBH criterion. This also goes to the other direction.