# state-space reduction

I am confused about state-space reduction. I learned it in the class but am not skilled in it.

If $A,B,C,D$ matrices are given with values, we can
1. find its controllability matrix to see if controllable, if uncontrollable
2. find a transformation matrix $P$
3. reduce the state space to a controllable state space representation

However, if NOT given values, just like the following:

Can I reduce the above state space representation to the following:

• Why would you give up those states? Why is their information negligible? – Max Herrmann Apr 21 '15 at 10:28
• Because of the uncontrollability and unobservability. I am not sure if it is so, and this is what I am confused about. – sleeve chen Apr 21 '15 at 18:43

I've found Lemma 3.20 in Zhou/Doyle's Robust and Optimal Control on p.72 which might be what you're looking for:

Lemma 3.20 Let $\left[\begin{array}{cc}A & B \\ C & D\end{array}\right]$ be a state space realization of a (not necessary stable) transfer matrix $G(s)$. Suppose there exists a symmetric matrix

$$P = P^* = \left[\begin{array}{cc}P_1 & 0 \\ 0 & 0\end{array}\right]$$

with $P_1$ nonsingular such that

$$AP + PA^* + BB^* = 0.$$

Now partition the realization $(A,B,C,D)$ compatibly with $P$ as

$$\left[\begin{array}{ccc}A_{11} & A_{12} & B_1 \\ A_{21} & A_{22} & B_2 \\ C_1 & C_2 & D \end{array}\right].$$

Then $$\left[\begin{array}{cc}A_{11} & B_1 \\ C_1 & D \end{array}\right]$$ is also a realization of $G$. Moreover, $(A_{11},B_1)$ is controllable if $A_{11}$ is stable.

$A^*$ denotes the complex conjugate of $A$ here.

• Nice, my question is exactly from the paper of Doyle's student. – sleeve chen Apr 22 '15 at 16:06