Dual of a MIMO control system

I am not sure to understand your question but I will start with writing this and I will update the answer based on comments.

First of all, the term "dual" used to talk about duality between controllability and observability is a bit misleading because it does not correspond, per se, to duality in the sense of an inner-product, even if there are connections between the two. From now on, I will refer to the system $\Sigma^T:=(A^T,C^T,B^T,D^T)$ to be the transposed system associated with $\Sigma:=(A,B,C,D)$. The transfer function, $G$, of $\Sigma$ is given by $G(s)=C(sI-A)^{-1}B+D$. Therefore, the transfer function of $\Sigma^T$ is just $G(s)^T=B^T(sI-A^T)^{-1}C^T+D^T$

Now, assume for simplicity that the system is stable, i.e. $A$ is Hurwitz stable, then the system $\Sigma$ maps real $L_2$ signals to real $L_2$ signals, and real $L_1$ signals to real $L_1$ signals.

As a result, we can look at the quantity $$ \langle z,\Sigma w\rangle:=\int_0^\infty z(s)^T\Sigma[w](s)ds. $$ Assuming that the signals are in both $L_1$ and $L_2$, we can use Plancherel theorem to obtain $$ \int_0^\infty z(s)^T\Sigma[w](s)ds=\int_{-\infty}^\infty \widehat{z}(j\omega)^*G(j\omega)\widehat{w}(j\omega)d\omega $$ Hence, $$ \int_0^\infty z(s)^T\Sigma[w](s)ds=\int_{-\infty}^\infty (G(j\omega)\widehat{z}(j\omega))^*\widehat{w}(j\omega)d\omega $$ and we get that $$ \int_0^\infty z(s)^T\Sigma[w](s)ds=\int_0^\infty \Sigma_d[z](s)^Tw(s)ds $$ where $\Sigma_d$ is the dual system of $\Sigma$ in the sense of this inner-product and the transfer function associated with this operator is $$ \begin{array}{rcl} G(j\omega)^*&=&B^*(j\omega I-A)^{-*}C^*+D^*\\ &=&B^T(-j\omega I-A^T)^{-1}C^T+D^T\\ &=&-B^T(j\omega I+A^T)^{-1}C^T+D^T \end{array} $$ which indicates that a state-space representation for $\Sigma_d$ is given by $(-A^T,-C^T, B^T,D^T)$.

We can make the following remarks:

• First, observe the negative sign, which was not present in the transposed system.
• Second, observe that if $\Sigma$ is stable, then $\Sigma_d$ is anti-stable (all the eigenvalues have positive real-part).
• Third, observe that $\Sigma$ controllable/observable if and only if $\Sigma_d$ observable/controllable (this comes from Kalman's conditions). So, we recover the standard results of the literature on the topic.

Define now the possible state-space representation for $\Sigma_d$ $$ \begin{array}{rcl} \dot{\tilde{x}}(t)&=&-A^T\tilde{x}(t)-C^T\tilde{w}(t)\\ \tilde{z}(t)&=&B^T\tilde{x}(t)+D^T\tilde{w}(t) \end{array} $$ and consider the change of variables for the time $t=-\tau$, and let $\bar{x}(\tau)=\tilde{x}(-t)$, $\bar{z}(\tau)=\tilde{z}(-t)$, and $\bar{w}(\tau)=\tilde{w}(-t)$. Then, $$ \begin{array}{rcl} \dot{\bar{x}}(\tau)&=&A^T\bar{x}(\tau)+C^T\bar{w}(\tau)\\ \bar{z}(\tau)&=&B^T\bar{x}(\tau)+D^T\bar{w}(\tau). \end{array} $$

Therefore, this shows that the transposed system is the dual system with time reversed.

What I have done is only valid for LTI system but the dual system for an LTV system $(A(t),B(t),C(t),D(t))$ would just be $$ \begin{array}{rcl} \dot{\tilde{x}}(t)&=&-A(t)^T\tilde{x}(t)-C(t)^T\tilde{w}(t)\\ \tilde{z}(t)&=&B(t)^T\tilde{x}(t)+D(t)^T\tilde{w}(t) \end{array} $$ but the time-reversal trick is not possible in this case.

Ok, I stop there and I will update based on comments and feedback.