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.
Best Answer
The first thing that should be corrected is the order of the transformation matrix, namely it should be
$$ V= \begin{bmatrix} V_{C\bar{O}} & V_{CO} & V_{\bar{C}\bar{O}} & V_{\bar{C}O} \end{bmatrix}. $$
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
$$ V_{C\bar{O}} = a_1\,u_1 + a_2\,u_2 = a_3\,u_3 + a_4\,u_4. $$