# Similarity Transformation misunderstanding

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 $$\|G(s)\|_{\infty}$$ $$G(s)=\left[\begin{array}{cc}{\frac{2}{s^{2}+0.2 s+1}} & {0} \\ {0} & {\frac{5}{s^{2}+0.4 s+4}}\end{array}\right]$$

• Could you add the actual matrix $P$ from the question? Sep 5, 2019 at 13:12
• 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. Sep 5, 2019 at 13:18
• 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)? Sep 5, 2019 at 13:26
• 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 $\sigma_3 = 0$. Sep 5, 2019 at 14:45
• Actually, more specifically: as long as it is possible to find such a $T$, the matrix $T = U^T$ will work. Sep 5, 2019 at 14:46

## 1 Answer

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.

• 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. Sep 5, 2019 at 17:26
• @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). Sep 5, 2019 at 17:31
• 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). Sep 5, 2019 at 17:33
• @Mr.Sh4nnon Then maybe it is a trick question, namely if $P$ is full rank then there doesn't exists such similarity transformation. Sep 5, 2019 at 17:38
• 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. Sep 5, 2019 at 17:49