showing any controllable system can be put in 'controller' form. I am looking at the proof of the following theorem: Let $\dot{x} =Ax + Bu$ be a controllable single input system, where $\Delta_A:= \det(\lambda I -A) = \lambda^n + a_1\lambda^{n-1} + \ldots + a_{n-1}\lambda + a_n$. Show this system is isomorphic to the system $\dot{x} = \bar{A}x + \bar{B}u$ where:
$$\bar{A} = \left(      \begin{matrix} 0 & 1 & 0& \dots & 0 & 0\\
0 &0&1 &\ldots &0&0 & \\
\vdots &\vdots &\vdots & & \vdots &\vdots &\\
0&0&0& \ldots & 0& 1\\
-a_n& -a_{n-1}& -a_{n-2}& \ldots & -a_2& -a_1
\end{matrix}\right) ,\quad\bar{b}=\left(  \begin{matrix} 0\\0\\ \vdots \\0\\0\\1 \end{matrix}\right)$$
This amounts to finding an invertible matrix $S$ such that $\bar{A}=S^{-1}AS$ and $\bar{B} = S^{-1}B$.
The proof however I really cannot follow. This theorem comes from a subject that I am taking for the second time. I distinctly remember begin quite impressed with this theorem last year when the teacher proved it with relative ease. However in the updated lecture notes we are given this year I cannot follow the proof at all. 
It starts of with defining the required matrix $S$ via $S = \left(q_1\quad q_2 \quad \ldots \quad q_n\right)$ where
\begin{eqnarray}
q_n&=&B\\
q_{n-1} &=& AB + a_{1}B\\
q_{n-1} &=& A^2B + a_{1}AB +a_{2}B\\
&\vdots&\\
q_1 &=& A^{n-1}B +a_{1}A^{n-2}B + \ldots +a_{n-1}B
\end{eqnarray}
Can anyone show me why this choice of $S$ works? Thanks a lot in advance for any help!

We need to show that $S \bar{A} = AS$. First look at $S \bar{A}$ (work it out).
$S \bar{A} = \begin{pmatrix} -a_n q_n & -a_{n-1} q_n + q_1 & -a_{n-2} q_n + q_2 & \dots & -a_1 q_n + q_{n-1} \end{pmatrix}$
Now $AS = \begin{pmatrix} A q_1 & A q_2 & \dots & A q_n \end{pmatrix}$. Let us look at each column seperately.
$A q_1 = A^n B + a_1 A^{n-1} B + \dots + a_{n-1} A B = -a_n B = -a_n q_n$.
It follows from the Cayley-Hamilton theorem, because $\Delta_A$ is the characteristic polynomial of $A$. Others will just follow from the definitions
$\begin{align}
A q_2 &= A^{n-1} B + a_1 A^{n-2} B + \dots + a_{n-2} A B = -a_{n-1} B + q_1 = -a_{n-1} q_n + q_1 \\
A q_3 &= A^{n-2} B + a_1 A^{n-3} B + \dots + a_{n-3} A B = -a_{n-2} B + q_2 = -a_{n-2} q_n + q_2 \\
&\vdots \\
A q_{n-1} &= A^2 B + a_1 AB = -a_2 B + q_{n-2} = -a_2 q_n + q_{n-2} \\
A q_n &= AB = -a_1 B + q_{n-1} = -a_1 q_n + q_{n-1} \\
\end{align}$
Hence, $S \bar{A} = AS$. Also, see that $S \bar{B} = q_n = B$.
Edit: Also note that $q_i$ are linearly dependant (ie. $S$ does not have an inverse) if the system is not controllable.
