# Inner product Proof,

So $V$ is an inner product space (finite dimensional) with inner product defined.

If $v$ and $w$ are vectors in $V$, how would one go about proving this?

$\langle \phi_\beta (x), \phi_\beta (y) \rangle ' = \langle [x]_\beta, [y]_\beta \rangle ' = \langle x, y \rangle$

The $\langle \rangle '$ is the standard inner product on $F^n$. $\beta$ is an orthonormal basis for V.

Attempt: It seemed to me that the first half of the equality would be obvious, because $\phi_\beta (x) =[x]_\beta$ and $\phi_\beta (y) =[y]_\beta$, or at least that is how I know $\phi_\beta$ to be defined. I don't know if that's rigorous though.

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Probably this is what you are looking for. If $\beta = \{e_1,e_2,...,e_n\}$ then for all $v \in V$, you have $v = \sum_i \langle v,e_i\rangle e_i$ and $[v]_\beta = (\langle v,e_1\rangle,\langle v,e_2\rangle,.....,\langle v,e_n\rangle) \in \mathbb{F}^n$ Hence immediately using $\langle e_i,e_j\rangle = \delta_{ij}$ you obtain $$\langle x,y\rangle = \sum_{i=1}^n \langle x,e_i\rangle\langle y,e_i\rangle = \langle [x]_\beta,[y]_\beta\rangle'$$
Thanks for your answer! But isn't $\beta = {e_1, ..., e_n}$ only one orthonormal basis? (i.e., the standard basis?) How would you show that for any orthonormal basis? Also, what about $\phi_\beta$?Is what I said above correct? –  user79449 May 26 '13 at 6:02
@user79449 $\{e_i\}$ may be any orthonormal basis, I have not used any property specific to standard basis, I don't know what do you mean by $\phi_\beta$ –  smiley06 May 26 '13 at 6:32
Ah, I see. $\phi_\beta$ is just a transformation representing a vector $x$ in $V$ in terms of the basis for $V$, which is why I was confused. –  user79449 May 26 '13 at 17:10