# Orthonormal basis and prove $\langle\phi_\beta(x), \phi_\beta(y)\rangle'=\langle[x]_\beta, [y]_\beta\rangle' = \langle x,y\rangle$

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?

• The inner product value is the same regardless of your choice of basis for the space. Proof by contradiction may be? – Vectorizer Mar 7 '20 at 18:24
• 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). – Marc van Leeuwen Mar 10 '20 at 13:30
• @MarcvanLeeuwen Sorry let me add up. – spruce Mar 10 '20 at 23:53

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, $$\langle [x]_\beta, [y]_\beta \rangle' = \langle x, y \rangle$$.