Let $A \in \mathbb{R}^{n\times n}, B \in \mathbb{R}^{n\times m}\,\, (n\geq m)$ and $$ A^\ast = \begin{pmatrix} A & -aI \\ a I & A \end{pmatrix} \, , \quad B^\ast= \begin{pmatrix} B & 0 \\ 0 & B \end{pmatrix} \, ,$$ where $a>0$.
According to the Popov-Belevitch-Hautus Eigenvector Test, $(A,B)$ is controllable iff no left eigenvector $x^\top$ of $A$ is orthogonal to the columns of $B$: $$ (A,B) \text{ controllable} \iff x^\top A = \lambda x^\top, \, x^\top B \neq 0 \, .$$
I would like to proof (and need some verification and/or guidance) that if $(A,B)$ is controllable and $A$ is diagonalizable, then $(A^\ast, B^\ast)$ is controllable.
My humble try of a proof:
Using the Kronecker product, we can write $$A^\ast = I_2 \otimes A + a \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \otimes I_n \, ,$$
where $I_n$ is the identity matrix of dimension $n$. Therefore we have for every left eigenvector $x^\top$ of $A$, two left eigenvectors of $A^\ast$, namely $$\begin{bmatrix}1 & \pm i \end{bmatrix} \otimes x^\top = \begin{bmatrix} x^\top & \pm i x^\top \end{bmatrix} \, . \quad (*)$$
Using the first assumption that $A$ is diagonalizable, all left eigenvectors of $A^\ast$ are of the form given in $(*)$. Employing the second assumption that $(A,B)$ is controllable, it follows that no left eigenvector of $A^\ast$ can be orthogonal to the columns of $B^\ast$ and so $(A^\ast, B^\ast)$ is controllable.
- Is this a valid proof?
- Any suggestions/improvements?
- Are there stronger results?