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The following question is from a System Theory test without answers or solutions:

Consider the continuous-time state-space representation

$\frac{d}{dt}x(t)=Ax(t)+Bu(t), \quad y(t)=Cx(t), \quad t\in \mathbb{R}^{+}$

With the marices given by

$A=\begin{bmatrix}0&-1&0&0\\-2&0&0&0\\1&0&-1&0\\0&0&0&-2 \end{bmatrix}, \quad B=\begin{bmatrix}0\\0\\-1\\1 \end{bmatrix} \quad \text{and} \quad C=\begin{bmatrix}1&0&0&-1 \end {bmatrix}$

Observe that the controllabilty and observabiltiy matrices are given by

$\mathcal{C}=\begin{bmatrix}0&0&0&0\\0&0&0&0\\-1&1&-1&1\\1&-2&4&-8 \end{bmatrix} \quad \text{and} \quad \mathcal{O}=\begin{bmatrix}1&0&0&-1\\0&-1&0&2\\2&0&0&-4\\0&-2&0&8 \end{bmatrix}$

Moreover, the reduced row-echelon forms of the controllability and observability matrices are given by

$\text{rref}(\mathcal{C})=\begin{bmatrix}1&0&-2&6\\0&1&-3&7\\0&0&0&0\\0&0&0&0 \end{bmatrix} \quad \text{and} \quad \text{rref}(\mathcal{O})=\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&0&0 \end{bmatrix}$

A Kalman decomposition of this state space realization with the matrices in the form

$\frac{d}{dt}z(t)=\bar{A}z(t)+\bar{B}u(t), \quad y=\bar{C}z(t), \quad t\in \mathbb{R}^{+}$

with $z(t)=M^{-1}x(t)$ can be obtained using a similarity transformation with the transformation matrix $M$ given by:

Answer:

$A) \quad M= \begin{bmatrix}0&0&1&0\\0&0&0&1\\1&0&0&0\\0&1&0&0 \end{bmatrix}$

$B) \quad M= \begin{bmatrix}0&0&1&1\\0&1&0&0\\-1&0&0&0\\1&1&1&1 \end{bmatrix}$

$C) \quad M= \begin{bmatrix}0&-1&1&2\\-2&0&1&-2\\1&0&1&2\\0&0&1&-2 \end{bmatrix}$

$D) \quad M= \begin{bmatrix}1&0&2&1\\0&0&-2&1\\2&1&2&1\\1&1&-2&1 \end{bmatrix}$

$E) \quad \text{None of the above}$

I am uncertain which answer is correct.

As far as I know the transformation matrix $M$ can be constructed as follows $M =\begin{bmatrix} V_{r\bar{o}}&V_{ro}&V_{\bar{r}\bar{o}}&V_{\bar{r}o} \end{bmatrix}$

in which the $V$ vectors are determined using the reachable space $\mathcal{R}$ and the unobservable space $\mathcal{UO}$ as follows:

$V_{r\bar{o}} \in \mathcal{R}$ and $V_{r\bar{o}} \in \mathcal{UO}$

$V_{ro} \in \mathcal{R}$ and $V_{ro} \notin \mathcal{UO}$

$V_{\bar{r}\bar{o}} \notin \mathcal{R}$ and $V_{\bar{r}\bar{o}} \in \mathcal{UO} $

$V_{\bar{r}o} \notin \mathcal{R}$ and $V_{\bar{r}\bar{o}} \notin \mathcal{UO}$

in which: $\mathcal{UO} =$ ker$(\mathcal{O}) = \begin{bmatrix}0\\0\\1\\0 \end{bmatrix} \quad$ and $\quad \mathcal{R} = \text{im}(\mathcal{C})= \begin{bmatrix}0&0\\0&0\\-1&1\\1&-2 \end{bmatrix}$

For answers $A$ and $B$ the third column ($V_{\bar{r}\bar{o}}$) is not a member of the set $\mathcal{UO}$. Which leaves only options $C$ and $D$. But as far as I know it is impossible to create the third columns of $C$ and $D$ using only $\mathcal{UO}$ and therefore $C$ and $D$ are also not correct. Thus the correct answer would be $E) \quad \text{None of the above}$.

Is there anyone with knowledge on the Kalman decompositon that can verify this approach?

Thanks in advance.

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You have to keep in mind that some of $V_{r\bar{o}}$, $V_{ro}$, $V_{\bar{r}\bar{o}}$ and $V_{\bar{r}o}$ can be empty. The only thing you know for sure is that the combined number of columns need to add up to the state dimension. For example a completely controllable and observable state space model would have $V_{ro}$ span the entire state space.

You do know that the span of $V_{r\bar{o}}$ and $V_{ro}$ should match the span of $\mathcal{R}$, which is of dimension 2. So the first two columns of $M$ should have the same span as $\mathcal{R}$. The only column of $\mathcal{UO}$ lies in the span of $\mathcal{R}$, so $V_{r\bar{o}}$ should have the same span as $\mathcal{UO}$. Since $\mathcal{UO}$ lied completely inside $\mathcal{R}$ means that $V_{\bar{r}\bar{o}}$ has to be empty. This would also imply that $V_{\bar{r}o}$ has two columns.

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