 Solution

$V^{\ast }$ is the space of linear functionals and the dual basis satisfies $v_i^{\ast }(v_j)={\delta }_{ij}$. Nondegeneracy and equal finite dimensions make $u\mapsto ⟨u,\cdot ⟩$ an isomorphism; evaluation similarly identifies $V$ with $V^{\ast \ast }$. Riesz representation applied to $v\mapsto ⟨Tv,u⟩$ gives the unique adjoint. If $T$ is diagonalizable, declare an eigenbasis orthonormal to make it self-adjoint. Conversely, for self-adjoint $T$ and invariant $W$, $⟨Tv,w⟩=⟨v,Tw⟩=0$ shows $W^{\perp }$ invariant.