I am working through Theorem 6 in Chapter 8 of Hoffman and Kunze. (The section is Linear Functionals and Adjoints.)
The statement is
Let $V$ be a finite-dimensional inner product space, and $f$ a linear functional on $V$. Then there exists a unique vector $\beta$ in $V$ such that $f(\alpha) = (\alpha | \beta)$ for all $\alpha$ in $V$.
(EDIT: The following is written in terms of a real vector space.)
I am fine with the fact that this essentially follows from what $f$ does to the elements of a basis for $V$, in other words:
For an arbitrary* $\mathcal{B} = \{e_1, \ldots, e_{n} \}$ (where $n = \dim V\;\;$) we can assign $\xi_{i} = f(e_{i})$ for $i=1,\ldots, n$. Then any $\alpha \in V$ can be written as $\alpha = x_{1}e_{1} + \cdots + x_{n}e_{n}$ so that $$\begin{align} f(\alpha) &= f(x_{1}e_{1} + \cdots + x_{n}e_{n}) \\ \\ &= x_{1}f(e_{1}) + \cdots + x_{n}f(e_{n}) \\ \\ &= x_{1} \xi_{1} + \cdots + x_{n}\xi_{n}.\end{align}$$
At this point we can identify the $\xi_{i}$'s as being independent of $\alpha \in V$, and only having been dependent on our choice of a basis. With this, we can simply assert that $\beta = (\xi_{1}, \ldots, \xi_{n})$ (and is in fact unique).
Now, this is apparently not the whole story! *According to H. and K., the coordinates of our fixed vector $\beta$ are determined by the action of $f$ on an orthonormal basis for $V$ rather than an arbitrary basis.
Question: What role does the orthogonality (or even the 'orthonormality') of the basis have in this argument? It seems to work just fine for an arbitrary basis as far as I can tell.