 Solution

A Hermitian matrix obeys $H^{†}=H$, where the dagger is conjugate transpose. If $Hv=\lambda v$ with $v\ne 0$, then $v^{†}Hv=\lambda v^{†}v$. This scalar is real because its complex conjugate equals $v^{†}{H}^{†}v=v^{†}Hv$. Since $v^{†}v>0$, the eigenvalue $\lambda$ is real.

For the Gram matrix, conjugate symmetry of the inner product givesThus $G$ is Hermitian. The assumption that the vectors form a basis additionally makes $G$ positive definite, though only Hermiticity is needed here.