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I have the following system:

$$ \dot x(t)=\begin{bmatrix} -2 & 1 & 2 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}x(t)+\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}u(t) $$

The controllability matrix is:

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

I have only been able to find the following rule so far: any $x\in\text{range}(P)$ is reachable. However, my problem is the following:

Given the initial state $x_0=\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$, is it possible to steer to $x_1=\begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix}$? How about to $x_2=\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}$?

I know that the answer is "yes" for $x_1$ but"no" for $x_2$. Interestingly, we have $x_1-x_0\in\text{range}(P)$ but $x_2-x_0\notin\text{range}(P)$ so it seems like we just need to check whether (final state minus initial condition) are in the range of the controllability matrix. Can anyone confirm this result and formally explain why this is so? Thanks a lot.

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For the specific system the uncontrollable dynamics are defined from $$z:=x_1-x_3=q^Tx$$ where $q=[\matrix{1\:0\:-1}]^T$. Note that $$\dot{z}=-2z$$ and therefore $$z(t)=e^{-2t}z(0)$$ So if you want to steer the system from $x_0$ to $x^*$ (with $x^*\neq x_0$) then there must exist some $c\in(0,1)$ such that $$q^Tx^*-cq^Tx_0=0$$

Since in your example $q^Tx_0=1$ and $q^Tx_1=1$, $q^Tx_2=0$ there is no control (except impulsive) that can steer $x_0$ to $x_1$ or $x_0$ to $x_2$.

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