= Solution
Use the convention that the <inner product> is conjugate-linear in its first argument and linear in its second, as in part (b). The <adjoint operator> is characterized by
$$
\boxed{\langle Tv,w\rangle=\langle v,T^*w\rangle
\qquad(v,w\in V).}
$$
Existence and uniqueness follow from the finite-dimensional representation of <linear functionals> by the <inner product>. More concretely, choose an <orthonormal basis> $e_i$ and let $a_{ij}=\langle e_i,Te_j\rangle$. Then
$$
\langle e_i,T^*e_j\rangle=\langle Te_i,e_j\rangle
=\overline{a_{ji}},
$$
which defines $T^*$ uniquely and verifies the required identity by expansion in that <basis>. Thus its <matrix> is the conjugate transpose, proved here rather than assumed. A <self-adjoint operator> satisfies $T^*=T$, equivalently $\langle Tv,w\rangle=\langle v,Tw\rangle$ for every pair of vectors.
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