Why is the 'controllable subspace' actually controllable?

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?

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.

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.