For the dynamic system $\dot x = Ax + Bu$

There's a saying that this system is controllable when $Ker(B) \in Ker(A)$, which means that $u$ have the control in every dimension of $x$.

I have no problem with this theorem when you have single input $u$, which means $u$ is just a number and $B$ is with $n\times1$ dimension.

But I cannot completely understand when $u$ is a vector. Look at this example:

$A = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 &0&0&0\\0 &0&0&0\\0 &0&0&0\\ \end{bmatrix}$ , $B = \begin{bmatrix} 0&0\\0&0\\0&0\\0&0\\\end{bmatrix}$

This is obvious uncontrollable. However, look at the null space:

$Ker(A)=span\{\begin{bmatrix} 0\\1\\0\\0\\ \end{bmatrix}, \begin{bmatrix} 0\\0\\1\\0\\ \end{bmatrix}, \begin{bmatrix} 0\\0\\0\\1\\ \end{bmatrix} \}$

$Ker(B)=span\{\begin{bmatrix} 1\\0 \end{bmatrix}, \begin{bmatrix} 0\\1 \end{bmatrix} \} $

Thus we have $Ker(B) \notin Ker(A)$, but this system is uncontrollable.

Can anyone help me with this contradictory?

Thank you very much and I appreciate all of your replies!

  • $\begingroup$ Do you mean $\text{Ker}(B) \subseteq \text{Ker}(A)$? Regardless, the dimension of $\text{Ker}(A)$ and $\text{Ker}(B)$ may not be compatible as in your example. However, we can deduce that $(A, B)$ is UNcontrollable if and only if there exists $0 \neq w \in \text{Ker}(\lambda I - A^T) \cap \text{Ker}(B^T)$ for all $\lambda \in \mathbb{C}$, from the PBH criteria. $\endgroup$ – obareey Jan 27 '14 at 17:39
  • $\begingroup$ Hi obareey, thank you very much! Your answer perfectly solves my question. I've never heard PBH theorem before. But using it can indeed avoid the incompatible matrix dimension. But one more question, can you explain PBH theorem in the intuitive sense of space? In my understanding, PBH says system is controllable when none of A's row eigenvectors is orthogonal to B. So is it correct to say that the space of A is spanned by A's row eigenvectors? $\endgroup$ – princeward Jan 28 '14 at 2:35
  • $\begingroup$ More question follows if what I've assumed just now is true: 1) Why the system space is spanned by A's row eigenvector rather than column eigenvector? 2) What happens if A can only be put into Jordan form but cannot be diagonalized (has a repeated eigenvalue whose geometric multiplicity is less than algebraic multiplicity)? In this case, is the system space less than n-dimension? Because A has less than n eigenvectors in this case. Thank you very much!!! $\endgroup$ – princeward Jan 28 '14 at 2:52

By system space if you mean "the space of reachable states", then it is the range space of the controllability matrix

$C=\begin{bmatrix}B & AB & A^2B & \dots & A^{n-1}B\end{bmatrix}$

It is relatively easy to see this for discrete time systems, but also not much difficult for continuous time systems. It follows from the Cayley-Hamilton Theorem for both cases.

If $C$ has full rank, this means we can reach any state in the space $\mathbb{R}^n$, hence the system is said to be "completely controllable" (or reachable, which is the same thing for continuous time case).

The left eigenvector condition follows from the fact that column space (or range) of a matrix is the orthogonal complement of its left null space, see this. It means that $\text{Im}C = \mathbb{R}^n$ is the same condition with $\text{Ker}C^T = 0$, which is equivalent to PBH theorem, see this for proof.

So, if $A$ has a left eigenvector $w$, which is orthogonal to all columns of $B$, then $w^T C=0$, which means we cannot reach the states in the direction of $w$, i.e. $w \notin \text{Im}C$. This also means that uncontrollable subspace is the space spanned by the left eigenvectors of $A$ that are orthogonal to the columns of $B$ and its orthogonal complement is the controllable subspace, which is the space spanned by the columns of $C$.

For these reasons, controllability condition is also equivalent to $\text{rank} \begin{bmatrix}\lambda I - A & B\end{bmatrix} = n$ for all $\lambda \in \mathbb{C}$. So, I would like to think that, whenever $\lambda I - A$ looses a rank for some $\lambda$ (which is an eigenvalue of $A$), it must be completed with one of the columns of $B$, so that we can drive the system to whichever state we want by using the input corresponding to that column. It cannot be done if the corresponding left eigenvector is orthogonal to the columns of $B$, because it means that we also cannot reach $w$ by using the columns of $B$, i.e. $w \notin \text{Im}B$ and $w \notin \text{Im}(\lambda I - A)$.

I hope this is the intuition you are looking for.

Note: These facts do not change if $A$ is not diagonalizable, as we are not interested in the vectors that are not left eigenvectors of $A$, because they are already in $\text{Im}(\lambda I - A)$ for all $\lambda \in \mathbb{C}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.