Let $\{v_1, v_2, \ldots, v_n\}$ be an orthonormal basis for a
finite-dimensional inner product space $V$ over some field $F$. For
any $x, y$ in $V$, $\langle x, y \rangle = \sum\limits_{i = 1}^n
\langle x, v_i \rangle \overline {\langle y, v_i \rangle}$. Prove that
if $\beta$ is an orthonormal basis for $V$ with inner product
$\langle\cdot, \cdot\rangle$, then for any $x,y \in V$,
$\langle\phi_\beta(x), \phi_\beta(y)\rangle'=\langle[x]_\beta,
[y]_\beta\rangle' = \langle x,y\rangle$, where $\langle \cdot, \cdot \rangle'$ denotes
the standard inner product.
What's the general idea in proving this?
Best Answer
This is an immediate consequence from Parseval's Identity, which you already stated. Recall that if $V$ is a nonzero finite-dimensional inner product space with orthonormal basis $\beta = \{ v_1, \dots, v_n \}$, then every $x \in V$ can be written as $$x = \sum_{i = 1}^n \langle x, v_i \rangle v_i.$$
Similar, $y \in V$ can be written as $y = \sum_{i = 1}^n \langle y, v_i \rangle v_i$. Here, $\langle x, v_i \rangle$ and $\langle y, v_i \rangle$ are the coordinates of $x$, respectively $y$, in this basis. Now you just have to observe that the right hand side of Parseval's Identity
$$\langle x, y \rangle = \sum_{i = 1}^n \langle x, v_i \rangle \overline{\langle y, v_i \rangle}$$
is the definition of the inner product in $F^n$. Hence, $\langle [x]_\beta, [y]_\beta \rangle' = \langle x, y \rangle$.