whether a Gramin matrics is symetric or not?

Question

A Gramin matrix is defined as $$ G^c(t_0,t_f)=\int_{t_0}^{t_f} \exp\bigl((t_0-t)A\bigr) BB^T \Bigl(\exp\bigl((t_0-t)A)\Bigr)^T \,dt$$ where ${}^T$ is for transpose of matrices. How can i prove it?

Answer 1

Note that writing $$ C(t) := \exp\bigl((t_0-t)A\bigr)B $$ your integrand is $I(t) := C(t)C(t)^T$, hence is symmetric as $$ I(t)^T = \bigl(C(t)C(t)^T\bigr)^T = C(t)^{TT}C(t)^T = C(t)C(t)^T= I(t) $$ So $$ G^c(t_0,t_f)^T = \left(\int_{t_0}^{t_f} I(t)\,dt\right)^T = \int_{t_0}^{t_f} I(t)^T\,dt = \int_{t_0}^{t_f} I(t)\, dt = G^c(t_0, t_f) $$ is symmetric.

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Addendum: $G^c(t_0, t_f)$ is allways positive semidefinite, as for any $\xi$: $$\xi^T I(t)\xi = \bigl(C(t)^T\xi\bigr)^T\bigl(C(t)^T\xi\bigr) \ge 0 $$