# Minimal State-Space-Realization using SVD

A similarity transformation with U produces so following state space model

$$$$(A, B, C, D) \stackrel{\tilde{x}=U^{\prime} \cdot x}{\longrightarrow}\left(U^{\prime} \cdot A \cdot U, U^{\prime} \cdot B, C \cdot U, D\right)=:(\tilde{A}, \tilde{B}, \tilde{C}, \tilde{D})$$$$

As far as I understand the same is possible with V for the observability case, since this formula above can be written in general with T which leads to the Kalman decomposition.

In my professors script I found this example:

$$$$A=\left[\begin{array}{cc}{-3} & {-2} \\ {1} & {0}\end{array}\right], B=\left[\begin{array}{c}{1} \\ {0}\end{array}\right], C=\left[\begin{array}{ll}{1} & {1}\end{array}\right], \quad D=0$$$$ $$$$\tilde{A}=V \cdot A \cdot V^{\prime}=\left[\begin{array}{cc}{-1} & {-3} \\ {0} & {-2}\end{array}\right], \tilde{C}=\left[\begin{array}{cc}{-1.414} & {0}\end{array}\right]$$$$

in which he than states:

Obviously, we have obtained the required form to separate the state vector into the observable and unobservable part. The first state is observable, the second is unobservable. The transfer function represenation of the reduced system is: $$$$g(s)=\frac{1}{s+2}$$$$

Do I assume right that there is a mistake? It should be $$\tilde{A}=V' \cdot A \cdot V$$ which would lead to $$A=\left[\begin{array}{cc}{-2} & {0} \\ {3} & {-1}\end{array}\right]$$

Then I could somehow see that the unobservable part falls away. The -2 in A remains which leads to the $g(s)$ mentioned.

Otherwise I don't get it. Why is the order of multiplication flipped?

EDIT:

So far I think my concerns are true. The rank of O(A,C) should not change after the transformation. It does when one uses VAV'. My proposed solution keeps the rank the same.