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Consider two following equivalent statements about the stabilizability:

\begin{equation} \dot{x} = Ax + Bu \\ \end{equation} where $x \in R^n$.

$1.$ The above system is stabilizable if and only if no eigenvector of $A^T$ with eigenvalue $Re\{\lambda\} \geq 0$ belongs to null($B^T$).

$2.$The above system is stabilizable if and only if $rank[A-\lambda I \;\;\; B] = n$ for $\lambda$ satisfying $Re\{\lambda\} \geq 0$.

Now consider the following $A$:

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

$A^T$ has one eigenvalue which is $0$ with a multiplicity of $3$. The eigenvectors of $A^T$then are:

$v = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$

Now, if I choose $B^T$ as:

$B^T$ = $ \begin{bmatrix} 1 & 1 & 1 \end{bmatrix}$

then $v$ is not in the null space of $B^T$ since $B^Tv \neq 0$. However if you check

$rank[A-\lambda I \;\;\; B] = rank[A \;\;\; B] = 2$

which should be $3$ since the statement $1$ is satisfied with the chosen $B$.

It seems there is a contradiction between the statements. Can you tell me what's wrong in here?

eet
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1 Answers1

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If a matrix has repeated eigenvalues then any non-zero linear combination of their obtained eigenvectors would also satisfy as eigenvector for that eigenvalue. So in your case for example

$$ u = v_1 - v_2 = \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} $$

also satisfies $A^\top u = \lambda\,u = 0$. However, $u$ does lie in the null space of $B^\top$.

So a more complete formulation for statement 1. would be:

A system is stabilizable if and only if for each eigenvalue $Re\{\lambda\}\geq0$ the intersect between the space spanned by the associated eigenvectors of $A^\top$ and null$(B^\top)$ is the origin.

Kwin van der Veen
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  • Does that mean we should check whether all the linear combinations of eigenvectors do not belong to the null space of $B^T$ (not just the eigenvectors)? In this example you found a linear combination which violates the criterion. – eet Jun 26 '22 at 10:59
  • I think if you have an eigenvalue with a multiplicity of $1$, then there will no problem. – eet Jun 26 '22 at 11:02
  • @eet For finding the intersection you could for example look at this question. And yes, when the multiplicity is one, then the statement in your question and the modified version in my answer yield the same result. – Kwin van der Veen Jun 26 '22 at 11:22
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    @eet A way to solve the problem is to consider the expression $\alpha VB=0$ where $V$ is a basis of the eigenspace of the considered eigenvalue and $\alpha\ne 0$ is a row vector you want to find. If there is no such $\alpha\ne0$, then the mode is controllable, otherwise it is not. In the current case, this yields $\alpha\begin{bmatrix}1\1\end{bmatrix}$, and equality holds with $\alpha=(1,-1)$. – KBS Jun 26 '22 at 20:11