Uncontrolable subset and stabilizability of a linear dynamical system I'm reading "Control System Design" by G. Goodwin, and I can't wrap my head around his definitions of uncontrolable subspace and stabilizability of a controlled dynamical system.
Consider a linear dynamical system of state $X\in\mathbb{R}^n$, controlled by an input vector $U\in\mathbb{R}^m$ : $$\dot{X}=AX+BU$$
We know that the controllable subset $R$ of the state space is the image of the controllability matrix $\mathfrak{C}$ : $$\mathfrak{C}=\begin{bmatrix}B&AB&..&A^{n-1}B\end{bmatrix}$$
And as such, its dimension is the rank $r$ of this matrix. Now let's assume this system isn't entirely controlable : $$r<n$$
Now, we can create a basis of this controllable subset $R$ by taking $r$ linearly independant columns $(v_1,..,v_r)$ of $\mathfrak{C}$ : $$R=\text{span}(v_1,..,v_r)$$
We now complete this basis of $R$ with $n-r$ vectors $(v_{r+1},..,v_n)$ of $\mathbb{R}^n$ in order to form a basis of $\mathbb{R}^n$. By doing this, we end up with a basis $(v_1,..,v_n)$ of $\mathbb{R}^{n}$ such that the $r$ first vectors $(v_1,..,v_r)$ of this basis span the controllable subset, and the $n-r$ next vectors $(v_{r+1},..,v_n)$ are necessarily uncontrolable directions of the state space.
Now, the author defines the uncontrollable subspace of the state space as the span of those last $n-r$ vectors, the ones that are necessarily uncontrolable directions : $$I=\text{span}(v_{r+1},..,v_n)$$
Now, to me this "uncontrollable space" is already a really strange notion. It doesn't include at all all the uncontrollable states of the state space, only an infinite fraction of them. And, this uncontrollable subspace depends on the arbitrary choice of the $n-r$ vectors used to complete the basis.
Like, if the state space is $\mathbb{R}^3$ (3D), and say $r=2$, the controllable subspace $R$ is a specific plane of $\mathbb{R}^3$ (no matter what pair of vectors you use as a basis for this plane, it remains the same plane). The set of uncontrollable states is then the complement of this plane in $\mathbb{R}^3$, which is not a linear subspace of $\mathbb{R}^3$. What Goodwin does, is taking an arbitrary line of $\mathbb{R}^3$ passing through the origin and not included in the controlable plane, and says "this is the uncontrolable subset $I$ of the state space".
I mean, yes we have $\mathbb{R}^3=R\oplus I$ (meaning that by summing elements of $R$ and $I$ we can reach all of $\mathbb{R}^3$), but beyond that this "uncontrolable subspace" is so arbitrary and limited that I can't see its utility.
And then comes the definition of stabilizability : this system is said to be stabilizable iff its uncontrolable subspace is stable, that is to say if there are no eigenvectors of $A$ associated with the unstable eigenvalues of $A$ (eigenvalues with a positive real parts) within the uncontrollable subset.
I can't wrap my head around it. Even if his uncontrolable subset is stable, that doesn't mean there isn't any unstable direction (eigenvector of $A$ associated with an unstable eigenvalue of $A$) that is outside of this subspace, and still isn't in the controllable subset (in my 3D example, another line of $\mathbb{R}^3$, that is neither the arbitrarily defined uncontrollable subset $I$ nor contained within the controllable plane). This is precisely because the author's definition of the uncontrollable subset is so limited and arbitrary.
I'd rather be enclined to think that a proper definition would more likely be : this system is said to be stabilizable iff all of its unstable directions are within the controllable plane (so that you can control of of the unstable modes of the system) - a definition that doesn't even require to define an uncontrollable subset.
The author then even goes further : if we transfer the representation of the system to the new basis, the new matrix $A$ takes has the following bloc structure : $$\overline{A}=\begin{bmatrix}A_{ctrl}&A_{1,2}\\0&A_{\textit{not-ctrl}}\end{bmatrix}$$
Which I agree with. He then says that the previous definition is equivalent to say that this system is stabilizable iff the eigenvalues of $A_{\textit{not-ctrl}}$ are stables. But what if the eigenvalues of $A_{\textit{not-ctrl}}$ are all stables, but $\overline{A}$ still has unstable eigenvalues that arent eigenvalues of $A_{\textit{ctrl}}$ neither ? To me, this feels like he implies that the eigen values of $\overline{A}$ (the same as those of $A$) are exactly the eigenvalues of $A_{ctrl}$ and those of $A_{\textit{not-ctrl}}$.
But juste like there are directions of $\mathbb{R}^3$ thare aren't the arbitrarily defined uncontrollable line $I$ nor contained within the controllable plane $R$, $\overline{A}$ can have eigenvalues of its own that aren't eigenvalues of neither $A_{ctrl}$ nor $A_{\textit{not-ctrl}}$, notably because there is a coupling bloc $A_{1,2}$.
I feel like I'm missing something obvious here, and I can't find it. Any possible explanation ?
 A: As you have said, the uncontrollable subspace forms a complement to the controllable subpace. In other words, $\Bbb R^n = R \oplus I$. That is, we can uniquely decompose every vector $x \in \Bbb R^n$ into the form $x = x_R + x_I$, with the "controllable component" $x_R \in R$ and "uncontrollable component" $x_I \in I$. A state $x \in \Bbb R^n$ is controllable if and only if its uncontrollable component is zero.
It is useful to have such a decomposition because the nature of the "state-update" matrix $A$ is completely determined by its behavior over these separate subspaces, since for any $x = x_R + x_I$, we have
$$
Ax = A(x_R + x_I) = Ax_R + Ax_I.
$$

Here is a continuous-time example. Suppose that we have
$$
A = \pmatrix{a_1 & 0\\0 & a_2},\quad  B = \pmatrix{1\\0}, \quad  C = \pmatrix{1&1}, \quad D = 0.
$$
It is easy to verify that our controllable subspace of $\Bbb R^2$ is the $x_1$-axis, i.e. the span of $(1,0)$. Any other one-dimensional subspace can be selected as the uncontrollable subspace, but it is convenient to take $I$ to be the span of $(0,1)$, since this space happens to be invariant under $A$ (note: such a complement is not always available).
Suppose that the initial state is given by $x(0) = (x_1,x_2)$. It is easy to see that for input $u(t)$, the state and output will be
$$
x(t) = \left(x_1 + e^{a_1t}\int_0^t e^{-a_1t}u(t)\,dt, \quad x_2 e^{a_2t}\right),
\\
y(t) = \left[x_1 + e^{a_1t}\int_0^t e^{-a_1t}u(t)\,dt\right] + x_2 e^{a_2t}.
$$
The first component of the sum, which corresponds to the controllable component of $x(t)$, can be stabilized with a suitable input. The second component, which corresponds to the uncontrollable component of $x(t)$, cannot be stabilized in this way. We could also say that the component $x_2 e^{a_2}t$ is itself an autonomous trajectory of the system: it transpires independently of the input.
We see from the above that the output is only stabilizable (i.e. can be "steered" so that $y(t) \to 0$) if $e^{a_2t} \to 0$.
Correspondingly, we see that $a_2$ is an eigenvalue of $A$ whose eigenvector $(0,1)$ is an element of the uncontrollable subspace $I$.

Suppose that we keep $v_1 = (1,0)$ as the basis for $R$, but instead take $v_2 = (1,1)$ as a basis for $I$. We find that
$$
Av_1 = a_1 v_1 + 0v_2, \\
A v_2 = \pmatrix{a_1\\a_2} = (a_1 - a_2)v_1 + a_2 v_2.
$$
So, the matrix of $A$ relative to the basis $\{v_1,v_2\}$ is
$$
\bar A = \pmatrix{a_1 & a_1 - a_2\\0 & a_2}.
$$
We indeed find that the eigenvalue $a_2$ of $A$ is associated with our uncontrollable subspace $I$. It is tricky, however, to figure out exactly what "associated with $I$" really means here.
One way to make sense of it is this. If we define the projection map $P_I(x_R + x_I) = x_I$, then we could say that the eigenvalue $\lambda$ of $A$ is "associated with $I$" if it is an eigenvalue of the map $T:I \to I$ defined by $T(x) = P_I(Ax)$.
