To show: $tr(CPC^{T}) = tr (B^{T}QB)$ with the help of Lyapunov equations $0 = AP + PA^{T} + BB^{T}$ and $0 = A^{T}Q + QA + C^{T}C$.
As shown in the text below; $||H||_2$ relates Lyapunov equation with the trace. But can anyone explain how the equation below comes into existence? Thanks
In order to show that $\mathrm{tr}(C\,P\,C^\top) = \mathrm{tr}(B^\top Q\,B)$ one can use a trace property of a product of matrices, which states that
$$ \mathrm{tr}(A^\top B) = \mathrm{tr}(A\,B^\top) = \mathrm{tr}(B^\top A) = \mathrm{tr}(B\,A^\top). $$
Namely, by defining the following matrix $M = C\,e^{A\,t} B$ and applying the stated trace property to $M$ it follows that
$$ \mathrm{tr}(M\,M^\top) = \mathrm{tr}(M^\top M), $$
and thus
$$ \mathrm{tr}\!\left(\underbrace{C\,e^{A\,t} B}_{M}\,\underbrace{B^\top e^{A^\top t} C^\top\!}_{M^\top}\right) = \mathrm{tr}\!\left(\underbrace{B^\top e^{A^\top t} C^\top\!}_{M^\top}\, \underbrace{C\,e^{A\,t} B}_{M}\right), $$
$$ \int_0^\infty \mathrm{tr}\!\left(C\,e^{A\,t} B\,B^\top e^{A^\top t} C^\top\right) dt = \int_0^\infty \mathrm{tr}\!\left(B^\top e^{A^\top t} C^\top\,C\,e^{A\,t} B\right)dt. $$