Adjoint of a linear map

Existence. Fix an orthonormal basis $(e_1,\ldots,e_n)$. For each $v$, the map $u\mapsto\langle f(u),v\rangle$ is linear in $u$, so there is a unique vector, call it $f^*(v)$, with $\langle f(u),v\rangle=\langle u,f^*(v)\rangle$ for all $u$.

Linearity of $f^*$. For $\alpha,\beta\in\mathbb{R}$ and $v,w$:
$$\langle u,f^*(\alpha v+\beta w)\rangle=\langle f(u),\alpha v+\beta w\rangle=\alpha\langle f(u),v\rangle+\beta\langle f(u),w\rangle.$$
By uniqueness, $f^*(\alpha v+\beta w)=\alpha f^*(v)+\beta f^*(w)$.

Matrix form. If $M$ is the matrix of $f$ in the orthonormal basis, then $f^*$ has matrix $M^T$ (transpose), because $\langle Mu,v\rangle=u^T M^T v=\langle u,M^T v\rangle$.

Conclusion. $f^*$ exists and is unique, called the adjoint of $f$. In matrix terms, $f^*$ corresponds to the transpose.