# Which constant matrices $A$ have the property that $\dot{x} =Ax+Bu$ is controllable for every non zero $B$?

Which constant matrices $A$ have the property that $\dot{x} =Ax+Bu$ is controllable for every non zero $B$?

Now to check controllability I usually check the rank of the matrix $$R = \left( B\ \ \ \ AB\ \ \ \ A^2B\ \ \ \ \ldots \ \ \ A^{n-1}B\right).$$ This rank has to equal $n$ where $n$ is the dimension of the $A$ matrix. Now if this matrix has to have rank $n$ for every non zero $B$ we can conclude something like $A$ has full rank or something or that $A$ spans $\Bbb R^n$ or something... I am sorry for not really having a solid idea on how to solve this... I tagged this linear algebra even though its a question from a systems control class because the solution probably needs only linear algebra.

Thanks for any help!

• This is only constant coefficient matrices? – Amzoti Nov 30 '13 at 13:57
• Yes, sorry ill include that! – Slugger Nov 30 '13 at 13:58

Hint. View $A$ as a complex matrix, despite all its entries are real. The pair $(A,B)$ is controllable if and only if $x^\ast B\ne0$ for every (possibly nonreal) left-eigenvector $x$ of $A$. Note that $x^\ast B=0$ if and only if $x$ is orthogonal to every column of $B$. So, we are looking for some matrices $A$ such that

$x^\ast b\ne0$ for every left-eigenvector $x$ of $A$ and for every nonzero real vector $b$.

Let us write $x=u+iv$ where $u,v$ are real vectors. Note that $x^\ast b=0$ if and only if $b$ is orthogonal to both $u$ and $v$. So, the problem not only involves the eigenspaces of $A$, but also has something to do with the size of $n$.

I don't think such a system matrix $A$ exists. Consider the Hautus criterion. A pair $(A,B)$ is controllable if and only if

$\text{rank} \begin{bmatrix} \lambda I - A & B \end{bmatrix} = n$

for any eigenvalue $\lambda$ of $A$. Without loss of generality assume that $A$ is in Jordan block form. Then, $\lambda I - A$ is a matrix whose at least one column and row is all zeros. But there is a non-zero $B$ such that the same row is all zeros, hence Hautus criterion fails.

For example, consider $A=\begin{bmatrix} \lambda & 1 \\ 0 & \lambda \end{bmatrix}$. Now, by Hautus criterion

$\text{rank} \begin{bmatrix} 0 & 1 & b_1 \\ 0 & 0 & b_2 \end{bmatrix}$

must be $n$. However there exists non-zero $B$ that this criterion fails. Hence, there is not a system matrix $A$ such that $(A,B)$ is controllable for any non-zero $B$.

However, note that if $\text{rank} B = n$, the system will always be controllable regardless of $A$.