As Marvis says in the comments, the problem reduces to showing that if $V$ is a finite-dimensional complex inner product space and $A : V \to V$ an operator such that $\langle v, Av \rangle = 0$ for all $v$, then $A = 0$. Letting $v$ run across all eigenvectors of $A$, the hypothesis implies that $A$ has all eigenvalues $0$, hence is nilpotent. Suppose by contradiction that $A \neq 0$, hence there exists a vector $v_0$ such that $v_1 = A v_0$ is nonzero. We may assume WLOG that $A v_1 = 0$, hence
$$\langle v_0 + v_1, A(v_0 + v_1) \rangle = \langle v_1, v_1 \rangle \neq 0$$
which is a contradiction. Hence $A = 0$.
Note that the corresponding assertion for real inner product spaces is false; the condition $\langle v, Av \rangle = 0$ is satisfied by all skew-symmetric matrices.