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If [A|B] is controllable, then $M_c=[B \quad AB \quad \dots \quad A^{n-1}B]$ has $n$ linearly independent colums. Then we consider the following table

$\begin{bmatrix} b_1& b_2& b_3 & \dots & b_m\\ Ab_1& Ab_2& Ab_3& \dots & Ab_m \\ \vdots & \vdots& \vdots & \vdots &\vdots\\ A^{n-1}b_1&A^{n-1}b_2&A^{n-1}b_3&\dots&A^{n-1}b_m \end{bmatrix}$

if we pick the vectors by going from the left to the right of the first row of the table, then the second row, and then the third row ..., and stop after $n$ linearly independent vectors are selected. Let the number of the selected vectors in the $j$th column is $k_j$. Then the bag $k = \{k1, k2, \dots,k_m\}$.

The Hermite indices are obtained by picking the vectors by going from the top to the bottom of the rst column of the table, then the second column, and then the third column ..., and stop after n linearly independent vectors are selected. Then we get $h = \{h1, h2, \dots,h_m\}$

Two questions on control theory

1)Why are the controllability indices,$k_j$, and the Hermite indices $h_j$, of [A|B] the same as those of [A + BF|B]? This is an exercise of Linear Systems CH4 3.21

2)Why the set of Hermite indices of [A|B], $h$, majorizes $k$, i.e. $k\prec h$

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2 Answers 2

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You can argue in terms of the Hautus test. Since

$$\begin{bmatrix}A+BF & B\end{bmatrix} = \begin{bmatrix}A & B\end{bmatrix} \begin{bmatrix}I &0 \\F & I\end{bmatrix} $$ and the rightmost matrix is full rank, the controllability properties are invariant.

The second is actually by definition or I didn't understand the question properly.

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  • $\begingroup$ Thanks! But what is controllability properties? And I know that the transformation $[P^{-1}AP|P^{-1}B]$ does not change the input-output relation of the system but why $[A B] P$ with $ P$ nonsingular maintains the controllability properties? $\endgroup$
    – Ye Xue
    Dec 5, 2017 at 15:42
  • $\begingroup$ @YeXue Because if $[\lambda I-A | B]$ is full rank for all $\lambda\in\mathbb{C}$ then the pair is controllable. And multiplication with a nonsingular matrix doesn't change the rank. $\endgroup$
    – percusse
    Dec 5, 2017 at 15:44
  • $\begingroup$ BTW I have modified my question. $\endgroup$
    – Ye Xue
    Dec 5, 2017 at 15:45
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The answers you want are in Section 5.7 of Linear Multivariable Control by WM Wonham, Theorems 5.9 and 5.10 particularly. Another version is Theorem 13.9 of Linear System Theory by WJ Rugh.

The proofs are somewhat involved. I don't think they can be dismissed as a simple exercise or direct consequence of controllability tests. I doubt you'll get a simpler or more concise explanation than those that can be found, for example, in either of the excellent references I suggest (I would like to be able to explain it in an easier way but I'm not capable of).

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