= Solution
Writing $u=\sum_i u_ie_i$ and $v=\sum_i v_ie_i$, evaluation against the <dual basis> gives
$$
\boxed{\phi(u,v)=\sum_i u_i\overline{v_i}.}
$$
\b[This has codomain $\mathbb C$, not the printed $\mathbb R$.] For a nonzero <vector space>, take $u=ie_1$ and $v=e_1$ to obtain $\phi(u,v)=i$. Only the diagonal values $\phi(v,v)$ must be real. The displayed formula defines the intended <complex inner product>.
It is linear in the first argument and conjugate-linear in the second:
$$
\phi(au+bw,v)=a\phi(u,v)+b\phi(w,v),\qquad
\phi(u,av+bw)=\overline a\phi(u,v)+\overline b\phi(u,w).
$$
It has <conjugate symmetry>, since $\phi(v,u)=\overline{\phi(u,v)}$. Finally,
$$
\phi(v,v)=\sum_i|v_i|^2\geq0,
$$
with equality exactly when every coordinate vanishes, equivalently $v=0$. This proves <positive definiteness>. Since $\phi(e_i,e_j)=\delta_{ij}$, the original <basis> is <orthonormal>. This is the <inner product determined by a basis>.
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