= Solution
Fix $y\in V$. The map $x\mapsto\theta(x,y)$ is a linear functional, so the finite-dimensional <Riesz representation theorem> gives a unique <vector> $\beta(y)$ such that
$$
\theta(x,y)=\langle x,\beta(y)\rangle
\qquad(x\in V).
$$
For <scalars> $\lambda,\mu$,
$$
\begin{aligned}
\langle x,\beta(\lambda y+\mu z)\rangle
&=\theta(x,\lambda y+\mu z)\\
&=\overline\lambda\,\theta(x,y)
+\overline\mu\,\theta(x,z)\\
&=\langle x,\lambda\beta(y)+\mu\beta(z)\rangle.
\end{aligned}
$$
Uniqueness gives
$$
\boxed{\beta(\lambda y+\mu z)=\lambda\beta(y)+\mu\beta(z)},
$$
so $\beta$ is linear.
For $\alpha\in\operatorname{End}(V)$, apply this result to
$$
\theta(x,y)=\langle\alpha x,y\rangle.
$$
It produces a unique <linear map> $\alpha^*$ satisfying
$$
\boxed{\langle\alpha x,y\rangle=\langle x,\alpha^*y\rangle},
$$
which proves existence and uniqueness of the adjoint.
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