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How can I go about proving that if $\|w\| = \|v\| = 1$, then $|\langle w, v\rangle| \leq 1$?

I know that $\langle w,v\rangle = w_1v_1 + w_2v_2 + \cdots $ but I am not sure how to use that to prove this.

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cauchy schwarz inequality –  Dominic Michaelis Mar 11 '13 at 15:53
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Hint: Recall the Cauchy-Schwarz Inequality.


$$|\langle x,y\rangle| ^2 \leq \langle x,x\rangle \cdot \langle y,y\rangle,$$

$$|\langle x,y\rangle| \leq \|x\| \cdot \|y\|.$$


You can explore its various applications and extensions at the Wikipedia Entry linked above.

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$\ddot\smile$ for my friend. –  B. S. Mar 12 '13 at 2:30
    
$\quad \ddot\smile\quad$ Good morning, Babak, dear friend! –  amWhy Mar 12 '13 at 2:34
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Remember the definition of cosine in an inner product linear space: if the angle between two vectors $\,u,v\,$ is $\,\theta\,$ , then

$$\cos\theta:=\frac{\langle\,u,\, v\,\rangle}{||u||\,||v||}$$

Now just remember that $\,|\cos\theta|\le 1\,$ ...(This follows from Cauchy-Schwarz).

If you defined the inner product by means of the above equation then trigonometry came first and you also know that $\,|\cos\theta|\le 1\,$, so either way...

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