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.
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
$$ V_{C\bar{O}} = a_1\,u_1 + a_2\,u_2 = a_3\,u_3 + a_4\,u_4. $$