# Kalman Decomposition of LTI System

I'm trying to find the Kalman decomposition of the following LTI system:

$$\dot{x}=Ax + bu, y=cx$$

where

$$A =\begin{pmatrix} -0.25 & 1.75 & 1.25 & 0.25 \\ 0.75 & -1.25 & -2.75 & -1.75 \\ -1 & 0 & -1 & -1 \\ 1.5 & 0.5 & 3.5 & 3.5 \end{pmatrix}$$

$$b=\begin{pmatrix} 1 & -1 & -3 & 3 \end{pmatrix}^T$$

$$c= \begin{pmatrix} 0.625 & -.875 & -1.125 & 0.375 \end{pmatrix}$$

but I keep getting stuck. The following is the method that I followed:

First, I found the controllability matrix $$C$$ and the observability matrix $$O$$, where I get

$$C= \begin{pmatrix} 1 & -5 & 9 & -13 \\ -1 & 5 & -9 & 13 \\ -3 & -1 & 5 & -9 \\ 3 & 1 & -5 & 9 \end{pmatrix}$$

and

$$O= \begin{pmatrix} 5/8 & -7/8 & -9/8 & 3/8 \\ 7/8 & 19/8 & 45/8 & 33/8 \\ 17/8 & 5/8 & 27/8 & 39/8 \\ 31/8 & 43/8 & 117/8 & 105/8 \end{pmatrix}$$

Then, I try to construct the transformation matrix $$V$$ s.t.

$$V= \begin{pmatrix} V_{CO} & V_{C\bar{O}} & V_{\bar{C}O} & V_{\bar{C}\bar{O}} \end{pmatrix}$$

(defined by my professor),

I find a basis $$u_1$$ and $$u_2$$ for the range of C (reachable space) as $$\begin{pmatrix}1\\-1\\-3\\3\end{pmatrix}$$ and $$\begin{pmatrix}-5\\5\\-1\\1\end{pmatrix}$$.

I find a basis $$u_3$$ and $$u_4$$ for the nullspace of O (unobservable space) as $$\begin{pmatrix}-1\\-2\\1\\0\end{pmatrix}$$ and $$\begin{pmatrix}-2\\-1\\-0\\1\end{pmatrix}$$.

To find a basis for $$C \cap \bar{O}$$, I can simply solve the equation

$$a_1u_1 + a_2u_2 = a_3u_3 + a_4u_4$$ and I get

$$\begin{pmatrix}a_1 \\ a_2 \\ a_3 \\ a_4 \end{pmatrix} = \begin{pmatrix}1 \\ 1 \\ -4 \\ 4 \end{pmatrix}=V_{C\bar{O}}$$

To get $$V_{CO}$$, I choose one of the bases for $$C$$. I chose $$V_{CO} = u_1$$

To get $$V_{\bar{C}\bar{O}}$$, I choose one of the basis vectors for $$\bar{O}$$. I chose $$V_{CO} = u_4$$

To get $$V_{\bar{C}O}$$, I can choose a linearly independent vector for $$R^4$$, and I chose $$V_{\bar{C}O} = \begin{pmatrix}0 \\ 0\\ 1\\ 0\end{pmatrix}$$

My choice of $$V = \begin{pmatrix} 1 & 1 & 0 & -2 \\ -1 & 1 & 0 & -1 \\ -3 & -4 & 1 & 0 \\ 3 & 4 & 0 & 1 \end{pmatrix}$$ doesn't transform A into the standard form that I expect.

Please help me in my understanding of how to obtain the correct transformation matrix $$V$$. My linear algebra skills are not the best; it's been a few years since I've taken a linear algebra course, so please correct any misunderstandings I have of how to perform the first steps of Kalman Decomposition.

$$V= \begin{bmatrix} V_{C\bar{O}} & V_{CO} & V_{\bar{C}\bar{O}} & V_{\bar{C}O} \end{bmatrix}.$$
The second correction is with regard to $$V_{C\bar{O}}$$, namely $$V_{C\bar{O}}$$ should both lie in the reachable space and the unobservable space. You did solve for correct gains, however you used those gains incorrectly since in order for $$V_{C\bar{O}}$$ to lie in both spaces it should be a linear combination of either spaces, so
$$V_{C\bar{O}} = a_1\,u_1 + a_2\,u_2 = a_3\,u_3 + a_4\,u_4.$$
• How is the second correction different than my attempt? I would still solve for $a_1$, $a_2$, $a_3$, and $a_4$ the same way, would I not? – Sean M Aug 1 '19 at 12:29
• @SeanM Yes, but the way you use $a_1$, $a_2$, $a_3$, and $a_4$ is different. – Kwin van der Veen Aug 1 '19 at 12:36
• Oh I understand now. So using the gains for the LHS, I calculate that $V_{C\bar{O}}$ = $\begin{pmatrix} -4 & 4 & -4 & 4 \end{pmatrix}^T$, which is the vector that my professor calculated. Thank you so much for your help – Sean M Aug 1 '19 at 12:42