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Suppose I have a discrete-time linear dynamical system $x_{t+1} = Ax_t + Bu_t$, with no constraints on $A$ and $B$, e.g. the system may not be controllable, A may not be stable. B may not be full column or row rank. Is there any relationship between the set of equilibrium points of the system (i.e. the $x^*$ such that there exists a $u^*$ where $x^* = Ax^* + Bu^*$) and the controllable subspace? e.g. is $x^*$ guaranteed to be in the system's controllable subspace? Or vice versa? If not, are there any (non-trivial, e.g. not "the system is controllable") conditions on $A$ and/or $B$ for which it's true that one is a subset of or equal to the other?

Edit (01/24/25): I think you're referring to a comment by @KBS on this question, which is partly why I am confused. If the set of equilibrium points is maintainable, I would expect them to also be controllable. But projecting an equilibrium point $x^*$ onto the controllable subspace gives me a different point $x^c$, which is not even an equilibrium point (i.e. trying to solve $x^c = Ax^c + Bu^c \implies u^c = B^+(I-A)x^c$ gives a $u^c$ for which $Bu^c \neq (I-A)x^c$). So it seems they are not the same, and I'm trying to see why, and what their intersection is (if any).

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    $\begingroup$ (1) if the system is not at least stabilizable, perturbations cause you to leave the equilibrium configuration. So you usually need that. (2) WLOG you can look at the origin since you can always rewrite your system relative to the target equilibrium config with the same $(A, B)$. Maybe you can clarify a bit more on what is driving this question. Because i think others may have already hinted at the answer in your other q. Otherwise maybe it can stand as is. The above isn't an answer, but worth thinking about. $\endgroup$ Commented Jan 24 at 6:25
  • $\begingroup$ Made a clarifying edit. But I'm confused about #2 -- I think there's some loss of generality... There can exist non-equilibrium points $x_{ne} \neq 0$ that don't have a corresponding steady-state input. However, $x=0$ is always an equilibrium with $u=0$. You're saying I re-center the system around $x_{ne}$ to effectively make it an equilibrium point. But this contradicts its non-equilibrium nature in the first place, because there is no corresponding shift that I can make on $u$. $\endgroup$ Commented Jan 24 at 20:09

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I will assume that the matrices are decomposed into controllable and uncontrollable blocks

$$ A = \begin{bmatrix} A_1 & A_3 \\ 0 & A_2 \end{bmatrix} , B = \begin{bmatrix} B_1 \\ 0\end{bmatrix}. \tag{1} $$

If the matrices are not of this form then one could obtain such form using the Kalman decomposition.

The equilibrium subspace are all vectors in the kernel of the matrix given below, i.e. $v\in\text{ker}(M)$ with

$$ v = \begin{bmatrix} u^* \\ x^* \end{bmatrix}, \\ M = \begin{bmatrix} B & A-I \end{bmatrix}. $$

Substituting $(1)$ in $M$ yields

$$ M= \begin{bmatrix} B_1 & A_1-I & A_3 \\ 0 & 0 & A_2-I \end{bmatrix}. $$

When $A_2-I$ is full rank then the uncontrollable subspace only has the origin as equilibrium. Therefore, in that case the equilibrium subspace of the full system will be a subset of the controllable subspace.

When $A_2-I$ is not full rank (i.e. 1 is an eigenvalue of $A_2$) then the uncontrollable subspace has other equilibria other than the origin. In which case the equilibrium subspace is not a subset of the controllable subspace.

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  • $\begingroup$ Perfect answer, thank you! Small follow-up: the Wikipedia page talks about decomposing into a "reachable" subsystem, not a "controllable" one. Are these equivalent? $\endgroup$ Commented Jan 26 at 15:44
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    $\begingroup$ @dkv for LTI continuous time models these two are equivalent, but for LTI discrete time models, as considered in your question, there a difference as discussed here. $\endgroup$ Commented Jan 26 at 16:25
  • $\begingroup$ So if there is a difference, how should I interpret your answer, since you consider the controllable/uncontrollable blocks, whereas the Kalman decomposition separates it into reachable/etc blocks? $\endgroup$ Commented Jan 26 at 20:18
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    $\begingroup$ @dkv the difference between controllability and reachability of discrete LTI systems occurs when $A$ is nilpotent. However, the uncontrollable subspace are related to all modes from $A_2$ that have an eigenvalue of 1, and thus this part of $A_2$ is not nilpotent. $\endgroup$ Commented Jan 26 at 21:17

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