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I was thinking about this proof of the cauchy-schwarz inequality,

I wanna show that $$|\langle u,v\rangle|\leq|u||v|$$.

We know that,

$$|\langle u,v\rangle| = ||u||v|\cos{\theta}|$$ where $\theta$ is the angle between $u$ and $v$.

But $-1 \leq \cos{\theta} \leq 1$, so, $$0\leq|\langle u,v\rangle|\leq|u||v|$$ where the equality holds if and only if $|\cos{\theta}|=1$, that is, $u$ and $v$ are linearly dependent.

This is correct? For which spaces? Only for $\mathbb{R^n}$?

Thanks :]

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    $\begingroup$ This reasoning is correct in $\mathbb{R}^n$ not necessarily in different Hilbert spaces there the inner product is not defined as you did. So not with an angle and cosine. The inequality holds though. $\endgroup$ – Kaladin Mar 7 '14 at 23:44
  • $\begingroup$ So it's correct in every space where the inner product is defined as ⟨u,v⟩=|u||v|cosθ, right? Thank you :] $\endgroup$ – zairhenrique Mar 7 '14 at 23:48
  • $\begingroup$ I think that I has to hold in any Hilbert space. Because $u,v$ span two dimensional space. So you can work in this two dimensional space. Thus it is sufficient to prove the inequality in 2d. $\endgroup$ – tom Mar 7 '14 at 23:48
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    $\begingroup$ Actually, this is the definition of angles of vectors for an inner product space. $\endgroup$ – Berci Mar 8 '14 at 0:08

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