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.


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 $$ \langle x,y\rangle = \sum_{i=1}^n \langle x,e_i\rangle\langle y,e_i\rangle = \langle [x]_\beta,[y]_\beta\rangle' $$

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  • $\begingroup$ 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? $\endgroup$ – user79449 May 26 '13 at 6:02
  • $\begingroup$ @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 $ $\endgroup$ – smiley06 May 26 '13 at 6:32
  • $\begingroup$ 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. $\endgroup$ – user79449 May 26 '13 at 17:10
  • $\begingroup$ @user79449 In that case, it is exactly the definition what you have written above. $\endgroup$ – smiley06 May 27 '13 at 8:52
  • $\begingroup$ @user79449 If you are convinced, do you mind ticking to accept this as an answer ? $\endgroup$ – smiley06 Jun 3 '13 at 17:23

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