Why is the 'controllable subspace' actually controllable?

Question

I am looking at the Kalman decomposition of a linear system into 'controllable' and 'uncontrollble' subspaces. The references I am using are these lecture notes and section 3.3 of 'Robust and Optimal Control' by Zhou and Doyle.

If we have a linear system governed by:

$\dot{x} = Ax + Bu$

Where $x$ is the system variable vector, $u$ is an input vector and $A,B$ are matrices of appropriate dimension, then we can perform a Kalman decomposition, changing the variables to $x' = Tx$, which gives an equation of the form:

$ \frac{dx'}{dt} = \begin{bmatrix} A_{11} & A_{12} \\ 0 & A_{22} \\ \end{bmatrix} x' + \begin{bmatrix} B_1 \\ 0 \\ \end{bmatrix} u$

We can now write $x' = \begin{bmatrix} x'_1 \\ x'_2 \\ \end{bmatrix} $ so:

$ \frac{d}{dt} \begin{bmatrix} x'_1 \\ x'_2 \\ \end{bmatrix} = \begin{bmatrix} A_{11} & A_{12} \\ 0 & A_{22} \\ \end{bmatrix} \begin{bmatrix} x'_1 \\ x'_2 \\ \end{bmatrix} + \begin{bmatrix} B_1 \\ 0 \\ \end{bmatrix} \begin{bmatrix} u_1 \\ u_2 \\ \end{bmatrix}$

The texts I am using now go on to say that the vector $x_1'$ is controllable and $x_2'$ is not.

It is not clear to me that $x_1'$ is controllable, since its evolution involves the uncontrollable variable $x_2'$ (multiplying out the first matrix gives $A_{11} x_1' + A_{12}x_2' + ...$) . The texts I am using go on to show that the pair ($A_{11}, B_1$) is controllable, whilst ignoring the $A_{12}$ term.

My question is this: why can we say that the variable $x_1'$ is controllable, given that the expression for its evolution involves $x_2'$, which is uncontrollable?

Answer 1

Suppose $A_{11}\in\mathbb R^{m\times m}$, $A_{12}\in\mathbb R^{m\times n}$, $A_{22}\in\mathbb R^{n\times n}$, and $B_1\in \mathbb R^{m\times p}$.

Let $A = \begin{bmatrix} A_{11} & A_{12} \\ 0 & A_{22} \\ \end{bmatrix} $ and $B = \begin{bmatrix} B_1 \\ 0 \\ \end{bmatrix}.$

To show that $x_1'$ is controllable, we need to prove that for every $x_0\in \mathbb R^{m+n}$ and $y\in \mathbb R^m$ and $t>0$, there exists a $u:[0,1]\rightarrow \mathbb R^p\ $ such that $x_i(t)=y_i$ when $1\leq i\leq m$ where $x$ solves the IVP

$x(0)=x_0$ and $ \frac{dx}{dt} = A x + B u.$

Assume $(A_{11}, B_1)$ is controllable.

Let $x_H$ be the solution to the IVP

$x_H(0)= x_0$ and $ \frac{dx_H}{dt} = A x_H.$

Let $z = x_H(t)$.

The fact $(A_{11}, B_1)$ is controllable implies there exists a $u^*:[0,t]\rightarrow \mathbb R^m$ such that the solution to the IVP $w:[0,t]\rightarrow \mathbb R^m$, $w(0)=0$, and $w\hskip{1pt}'= A_{11} w + B_1 u^*$ satisfies $w_i(t) = y_i-z_i$ when $1\leq i \leq m$.

Let $$x=x_H + \begin{bmatrix} w \\ 0 \\ \end{bmatrix}. $$

A bit of algebra shows that $x'=A x + B u^*$, $x(0)= x_0$, and $x_i(t)=y_i$ when $1\leq i\leq m$, thus $x_1'$ is controllable.

I guess that you can summarize the above reasoning by saying that the fact that $(A_{11}, B_1)$ is controllable implies that you can find a $u^*$ that can control the first $m$ coordinates of $x$. You can use that control to nullify the influence of the remaining $n$ coordinates and simultaneously steer the first $m$ coordinates to any chosen values.

Answer 2

Because the $A_{12}$ term is irrelevant for controllability of the $x_1'$ states. To see this write the equation for $x_1'$ $$ \begin{align} \dot{x}_1'(t) &= A_{11} x_1'(t) + A_{12} x_2'(t) + B_1 u_1(t) \tag{1} \\ &= A_{11} x_1'(t) + A_{12} e^{A_{22} t} x_2'(0) + B_1 u_1(t) \end{align}$$ If $x_2'(0) = 0$ then it is obvious. But we can also solve this equation as follows: $$ x_1'(t) = e^{A_{11} t} x_1'(0) + \int_0^t e^{A_{11} (t - \tau)} A_{12} e^{A_{22} \tau} x_2'(0) d\tau + \int_0^t e^{A_{11} (t - \tau)} B_1 u_1(\tau) d\tau $$ Now, to reach an arbitrary final state $x_f$ at time $t_f$, we can select $u_1$ as $$ u_1(t) = B_1^T e^{A_{11}^T (t_f - t)} W_c^{-1} (t_f) \left( x_f - e^{A_{11} t_f} x_1'(0) - \int_0^{t_f} e^{A_{11} (t_f - \tau)} A_{12} e^{A_{22} \tau} x_2'(0) d\tau \right)$$ where $$ W_c (t) = \int_0^t e^{A_{11} (t - \tau)} B_1 B_1^T e^{A_{11}^T (t - \tau)} d\tau $$ is the controllability gramian.

So the system $(1)$ is controllable if and only if $(A_{11}, B_1)$ is controllable. Basically we can cancel out the parts that comes from the initial conditions. This means we can select $x(0) = 0$ without losing generality to obtain the results.