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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!

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    $\begingroup$ You need to change the order of the scalars. See the post below. $\endgroup$ – obareey Dec 6 '13 at 17:45
  • $\begingroup$ Thanks for your comment, I think I made the adequate changes to the definition of the $a_i$'s. $\endgroup$ – Slugger Dec 6 '13 at 17:56
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    $\begingroup$ Also $\bar{A} = S^{-1} A S$ $\endgroup$ – obareey Dec 6 '13 at 18:16
  • $\begingroup$ Ah yes thanks for that! $\endgroup$ – Slugger Dec 6 '13 at 19:36
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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.

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Here is another view (inspired by Luenberger's 1967 paper http://www.stanford.edu/dept/MSandE/cgi-bin/people/faculty/luenberger/pdfs/cfflms.pdf, this is a slightly different solution for a slightly simpler problem):

(Note that even Luenberger wrote that this "is somewhat more difficult to derive".)

Let $P = \begin{bmatrix} b & Ab & \cdots & A^{n-1} b \end{bmatrix}$. Since the system is cc., $P$ is invertible. It is straightforward to show that using the basis $P$, the system $(A,b)$ has the form $\left( \begin{bmatrix} 0 & 0 & 0& \dots & 0 & -a_n\\ 1 &0&0 &\ldots &0& -a_{n-1} & \\ \vdots &\vdots &\vdots & & \vdots &\vdots &\\ 0&0&0& \ldots & 0& -a_2 \\ 0& 0& 0& 0 & 1& -a_1 \end{bmatrix}, e_1 \right)$, where $e_k$ is all zeros except for a one in the $k$th position.

This is close, but we need the transpose instead, and the '$b$' matrix needs to be $e_n$ instead.

The key claim is that the system $(A^T, P^{-T} e_n)$ is cc. Let $W^T = \begin{bmatrix} e_n^T P^{-1} \\ \vdots \\ e_n^T P^{-1} A^{n-1} \end{bmatrix}$ (the transpose of the usual). We need to show that $W$ is invertible. Note that $e_n^T P^{-1} A^{i-1}A^{j-1}b = e_n^T P^{-1} P e_{i+j-1} = \delta_{n,i+j-1}$ for $i,j$ such that $2 \le i+j \le n$. Hence $W^TP =\begin{bmatrix} 0 & 0 & 0& \dots & 0 & 1 \\ 0 &0&0 &\ldots &1& * & \\ \vdots &\vdots &\vdots & & \vdots &\vdots &\\ 0&1& * & \ldots & * & * \\ 1& *& *& \ldots & *& * \end{bmatrix}$, where the $*$ entries do not matter in the context of showing that $W$ is invertible. Since $P$ and the right hand side are invertible, it follows that $W^T$ is invertible and hence $(A^T, P^{-T} e_n)$ is cc. Note that this computation also shows that $W^T P e_1 = e_n$.

Now we note that in the basis $W$, the matrix $W^{-1}A^TW$ has the form given above, hence the matrix $W^T A W^{-T}$ has the desired form. Furthermore, $W^Tb = W^T P e_1 = e_n$, as desired.

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The two representations $(A, \mathbf{b})$ and $(\bar{A}, \bar{\mathbf{b}})$ are isomorphic iff there is a bijection $\Phi:\mathbb{R^n} \rightarrow \mathbb{R^n}$ from the original states $\mathbf{x}(t)$ to the states $\mathbf{z}(t)$ in normal form: $\Phi(\mathbf{x}(t)) = W\mathbf{x}(t) = \mathbf{z}(t)$ where $W \in \mathbb{R}^{n \times n}$.

The linear transformation yields the following set of equations (argument $t$ omitted for sake of readability): $$ \begin{align} \mathbf{w}_1^T\mathbf{x} = z_1 \\ \mathbf{w}_1^T\dot{\mathbf{x}} = \mathbf{w}_1^T (A\mathbf{x} + \mathbf{b}u) = \mathbf{w}_1^TA\mathbf{x} + \underbrace{\mathbf{w}_1^T\mathbf{b}}_{=0}u = \dot{z}_1 = z_2 \\ \mathbf{w}_1^TA\dot{\mathbf{x}} = \mathbf{w}_1^TA (A\mathbf{x} + \mathbf{b}u) = \mathbf{w}_1^TA^2\mathbf{x} + \overbrace{\underbrace{\mathbf{w}_1^TA}_{\mathbf{w}_2^T}\mathbf{b}}^{=0}u = \dot{z}_2 = z_3 \\ \vdots \\ \mathbf{w}_1^TA^{n-1}\mathbf{x} = z_n. \end{align} $$ Matrixwise this reads $$ \underbrace{ \left[ \begin{array}{c} \mathbf{w}_1^T \\ \mathbf{w}_1^TA \\ \mathbf{w}_1^TA^2 \\ \vdots \\ \mathbf{w}_1^TA^{n-1} \end{array} \right]}_{=W}\mathbf{x} = \mathbf{z}. $$ Where the conditional equation for $\mathbf{w}_1^T$ is $$ \left[ \mathbf{w}_1^T\mathbf{b}, \mathbf{w}_2^T\mathbf{b}, ..., \mathbf{w}_n^T\mathbf{b} \right] = \left[ \mathbf{w}_1^T\mathbf{b}, \mathbf{w}_1^TA\mathbf{b}, ..., \mathbf{w}_1^TA^{n-1}\mathbf{b} \right], $$ which, from $\bar{\mathbf{b}} = [0 ... 0 \: 1]^T$, returns $$ [0 ... 0 \: 1] = \mathbf{w}_1^T \underbrace{[\mathbf{b} \: A\mathbf{b} \: ... \: A^{n-1}\mathbf{b}]}_{=:S(A, \mathbf{b})}. $$ If $S(A, \mathbf{b})$ is invertible $$\mathbf{w}_1^T = [0 ... 0 \: 1] \: S^{-1}(A, \mathbf{b}), $$ which is a unique solution and a necessary condition for $W$ to be invertible and $\Phi$ to be 1-1 and onto.

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