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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?

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    $\begingroup$ The inner product value is the same regardless of your choice of basis for the space. Proof by contradiction may be? $\endgroup$ – Vectorizer Mar 7 '20 at 18:24
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    $\begingroup$ You forgot to introduce $\phi_\beta$, and $[x]_\beta$ as well (my guess the latter is the $n$-tuple of coordinates of $x$ with respect to the basis $\beta$, but for $\phi_\beta$ I am at a loss). $\endgroup$ – Marc van Leeuwen Mar 10 '20 at 13:30
  • $\begingroup$ @MarcvanLeeuwen Sorry let me add up. $\endgroup$ – spruce Mar 10 '20 at 23:53
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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, $\langle [x]_\beta, [y]_\beta \rangle' = \langle x, y \rangle$.

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