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All of my scripts and textbooks about control theory introduce the similarity transformation. It is needed to work in the state space. However, none of them have any examples. I don't get what they are doing.

Then this exercise came without solution, which perfectly shows that I don't get it:

By means of a similarity transoformation P can be transformed into $\tilde{P}=\begin{bmatrix}\tilde{P}_1 \\ 0\end{bmatrix}.$Determine the necessary state transformation matrix T.

So a similarity transformation is $\tilde{P}=TP$. I search for T. My first problem is that I don't see what $\tilde{P}_1$ is. In my view P is a matrix of undefined size. Therefore the 1 is an index. But I'd need two indexes for a matrix element. So my assumption is probably wrong.

My second problem is how to get T. If I know what $\tilde{P}_1$ is I could do something like $T=\tilde{P}_1P^-1$. But then I also need to know what P is or how else am I supposed to invert it?

This exercise should be rather simple since it gave only 1 point. I hope it fits the requirements for Mathematics StackExchange and is not too trivial. I am glad for any advice which points me in the right direction of similarity transformations in general.

EDIT: Apparently some matrix P is needed. There is never mentioned that it is this P, but in a previous exercise P is mentioned as:

The controllability matrix $P=\left[\begin{array}{lll}{A^{2} B} & {A B} & {B}\end{array}\right]$ of a state-space model $(A,B,C,D)$ has the following SVD. Determine the kernel and image space of $P$. $$ P=\left[\begin{array}{lll}{u_{1}} & {u_{2}} & {u_{3}}\end{array}\right]\left[\begin{array}{ccc}{\sigma_{1}} & {0} & {0} \\ {0} & {\sigma_{2}} & {0} \\ {0} & {0} & {\sigma_{3}}\end{array}\right]\left[\begin{array}{c}{v_{1}^{T}} \\ {v_{2}^{T}} \\ {v_{3}^{T}}\end{array}\right] $$

EDIT 2: Since there seems that some crucial information is missing I will add the first exercises as well. I don't think it contains information that will help, but I'll give it a try. I strongly assumed those exercises are independent. What's following is exercises 1.a) so it actually came first. The first edit was 1.b) and what I actually asked is in 1. c)

Each element of the following transfer function G(s) has the bode diagram shown in the succeeding bode plots. Estimate $\|G(s)\|_{\infty} $ $$ G(s)=\left[\begin{array}{cc}{\frac{2}{s^{2}+0.2 s+1}} & {0} \\ {0} & {\frac{5}{s^{2}+0.4 s+4}}\end{array}\right] $$

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    $\begingroup$ Could you add the actual matrix $P$ from the question? $\endgroup$ Sep 5, 2019 at 13:12
  • $\begingroup$ Well, so I am not the only one thinking there should be a matrix? The exercise has the number 1. c). There is however another P in the exercise b). It is never mentioned that those are related but I assume thats it. Still no numbers. I'll add it. $\endgroup$
    – Mr.Sh4nnon
    Sep 5, 2019 at 13:18
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    $\begingroup$ Are we given that $\sigma_3 = 0$? Do we know anything about the $\sigma_i$? Is there any information we can use from 1. a)? $\endgroup$ Sep 5, 2019 at 13:26
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    $\begingroup$ I agree, finding $A$ and $B$ seems like too much for a one point question. It seems like you're supposed to find a matrix $T$ for which $TP$ has a row of zeros, but this should only be possible if $P$ does not have full rank, in other words only if $\sigma_3 = 0$. $\endgroup$ Sep 5, 2019 at 14:45
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    $\begingroup$ Actually, more specifically: as long as it is possible to find such a $T$, the matrix $T = U^T$ will work. $\endgroup$ Sep 5, 2019 at 14:46

1 Answer 1

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A similarity transformation for a state space model $(A,B,C,D)$ can be described with

\begin{align} \hat{A} &= T\,A\,T^{-1}, \\ \hat{B} &= T\,B, \\ \hat{C} &= C\,T^{-1}, \\ \hat{D} &= D. \end{align}

So the transformed controllability matrix would be

\begin{align} \tilde{P} &= \begin{bmatrix} \hat{A}^2 \hat{B} & \hat{A}\,\hat{B} & \hat{B} \end{bmatrix} \\ &= T \begin{bmatrix} A^2 B & A\,B & B \end{bmatrix} \\ &= T\,P. \end{align}

It can be noted that $T$ should always be invertible (otherwise $\hat{A}$ and $\hat{C}$ can not be calculated). From this it also follows that the rank of $P$ should equal the rank of $\tilde{P}$. Now since $\tilde{P}$ should contain a row of zeros means that it can't be full rank so neither can $P$. This would mean that either one or more of the three $\sigma$ should be zero or $u_i\in\mathbb{R}^n$ with $n>3$. However in the previous exercise you had to calculate the kernel of $P$ so I assume that one of these requirements would be satisfied. From this you can also deduce that $T$ could be obtained using

$$ T = \begin{bmatrix} T_1 \\ T_2 \end{bmatrix}, $$

with $T_2$ the left null space of $P$ and $T_2$ such that $T$ is square and full rank.

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  • $\begingroup$ Thank you for the answer. It is very informative. But I still can't see it. [T1;T2] is supposed to be the solution? Thats it? Left null space wasn't even discussed in this course so is there another way one can come to this solution? I still don't see what P1 is, where it came from and how it was used. $\endgroup$
    – Mr.Sh4nnon
    Sep 5, 2019 at 17:26
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    $\begingroup$ @Mr.Sh4nnon How did you answer 1b) because the information given in your question doesn't seems to be enough to answer that, because I suspect that that information can also be used for 1c). $\endgroup$ Sep 5, 2019 at 17:31
  • $\begingroup$ Actually I did not. math.stackexchange.com/questions/3344081/… here we discuessed 1. b). I wrote down now everything I've got. I only left the Bode plots but I wrote down G(s). $\endgroup$
    – Mr.Sh4nnon
    Sep 5, 2019 at 17:33
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    $\begingroup$ @Mr.Sh4nnon Then maybe it is a trick question, namely if $P$ is full rank then there doesn't exists such similarity transformation. $\endgroup$ Sep 5, 2019 at 17:38
  • $\begingroup$ Well, fu... thank you very much. I hope you are right. If he would at least write what P is then you could argue that some one knowing linear algebra might get this. But without that it's just a joke. $\endgroup$
    – Mr.Sh4nnon
    Sep 5, 2019 at 17:49

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