# prove/show algebraic equivalence of 2 3x3 systems.

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.

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: $$AT_i=T_i \Lambda$$