Role of orthonormal basis for: Linear functional on a finite dimensional inner product space is 'inner product with a fixed vector in the space'

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.

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1 Answer

If the basis $\{e_1,\ldots,e_n\}$ is not orthonormal, then the inner product of the vectors $\alpha = x_1e_1 + \cdots + x_n e_n$ and $\beta = \xi_1 e_1 + \cdots + \xi_n e_n$ is not equal to $x_1\xi_1+\cdots+x_n\xi_n$. Thus, while it is true that $f(\alpha) = x_1\xi_1+\cdots+x_n\xi_n$, it is not true that $f(\alpha)=(\alpha|\beta)$.

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Jim, thank you for pointing this out. Now I recognize that without orthogonality of the basis, $\alpha \neq (x_{1},\ldots,x_{n})$, and without unit length of each basis element, the products $\| e_{i}\|x_{i}\xi_{i} \neq x_{i}\xi_{i}$. – Tom Stephens Sep 5 '10 at 4:21
I have to unaccept this answer because I am no longer convinced of the claim that $f(\alpha)\neq(\alpha | \beta)$ in the absence of an orthonormal basis. Could you expand on that a bit? – Tom Stephens Sep 7 '10 at 20:18
In general, $(\alpha|\beta) = \sum_{i=1}^n \sum_{j=1}^n x_i\xi_j(e_i|e_j)$ by the linearity of the inner product. If the basis is orthonormal, then $(e_i|e_j)$ is 1 for $i=j$ and 0 otherwise, in which case the sum reduces to the usual formula for the inner product. – Jim Belk Sep 8 '10 at 5:10