= Solution
A <Hermitian matrix> obeys $H^\dagger=H$, where the dagger is conjugate transpose. If $Hv=\lambda v$ with $v\ne0$, then $v^\dagger Hv=\lambda v^\dagger v$. This scalar is real because its complex conjugate equals $v^\dagger H^\dagger v=v^\dagger Hv$. Since $v^\dagger v>0$, the <eigenvalue> $\lambda$ is real.
For the <Gram matrix>, conjugate symmetry of the <inner product> gives
$$
(G^\dagger)_{ij}=\overline{G_{ji}}=\overline{\langle u_j,u_i\rangle}=\langle u_i,u_j\rangle=G_{ij}.
$$
Thus $G$ is <Hermitian>. The assumption that the vectors form a basis additionally makes $G$ positive definite, though only Hermiticity is needed here.
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