Consider the following two continuous-time state-space representations of the form
$\frac{d}{dt}x(t) = Ax(t)+Bu(t), \quad y(t)=Cx(t), \quad t \in \mathbb{R}^+$
With their matrices given by
$1) \quad \left[ \begin{array}{l|l}
A&B\\
\hline
C
\end{array} \right] = \left[ \begin{array}{lll|l}
3&0&0&0\\
0&1&-1&0\\
0&0&2&1\\
\hline
0&1&2
\end{array} \right] \quad 2)\quad \left[ \begin{array}{l|l}
A&B\\
\hline
C
\end{array} \right] = \left[ \begin{array}{lll|l}
1&0&0&1\\
0&2&-1&1\\
0&0&3&0\\
\hline
1&1&1
\end{array} \right]$
Note that for these systems $1$ and $2$ there exist invertible matrices $T_1$ and $T_2$, respectively, such that $AT_i=T_i\Lambda, \ \bar{B}\ \text{and} \ \bar{C}T_i, \ i=1,2$ with
$\Lambda = \begin{bmatrix}1&0&0\\0&2&0\\0&0&3\end{bmatrix}, \quad \bar{B}=\begin{bmatrix}1\\1\\0\end{bmatrix} \quad \text{and} \quad \bar{C}=\begin{bmatrix}1&1&0\end{bmatrix}$
Systems $1$ and $2$ are algebraically equivalent according to the answers. But i dont know how to check/prove this.
Normally I would prove algebraic equivalence by showing that:
$A_2=TA_1T^{-1}$ in which $T = \mathcal{C_1}\mathcal{C_2}^{-1}$ with $\mathcal{C_i}$ being the corresponding controllability matrix. This has always worked but now $\mathcal{C_2}$ is singular. So I don't know how to check/prove it.
Best Answer
Two state space realizations are always algebraically equivalent as long as the matrices have the same state dimensions and there exist invertible matrices $T_1$ and $T_2$, respectively for which the following holds: $AT_i=T_i \Lambda$