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Cauchy-Schwarz Inequality:

If $\textbf{u}$ and $\textbf{v}$ are vectors in a real inner product space $V$, then $$|\left\langle\textbf{u},\textbf{v}\right\rangle|\leq||\textbf{u}||\ ||\textbf{v}||$$

What will happen with the Cauchy-Schwarz inequality if the angle between the two vectors is zero?

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    $\begingroup$ Algebraically, what does it mean for the angle between $u$ and $v$ to be $0$? Answer this and the answer to your question will present itself. $\endgroup$ – Git Gud Nov 11 '14 at 16:37
  • $\begingroup$ @GitGud $|<\textbf{u},\textbf{v}>| = ||\textbf{u}|| ||\textbf{v}||$ ? I thought this, but it seemed a bit simple, considering the space left open in the past question paper in which one could answer it :) $\endgroup$ – user860374 Nov 11 '14 at 16:39
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the definition of the angle $\alpha\in[0,\pi]$ between $u,v$ is: $$ \cos\alpha = \frac{\langle u,v\rangle}{\|u\| \|v\|} $$

so $\alpha =0 $ iff $$ 1=\frac{\langle u,v\rangle}{\|u\| \|v\|}\iff \langle u,v\rangle = \|u\| \|v\| $$

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When angle between $v,w$ is $0$ (in other words $v=\alpha w$ or $w=\alpha v$ for some $\alpha$), then (if $w = \alpha v$):

$$|\langle w,v\rangle|=|\langle \alpha v,v\rangle|=|\alpha\langle v,v\rangle|=|\alpha|\|v\|^2=|\alpha|\|v\|\|v\|=\|\alpha v\|\|v\|=\|w\|\|v\|$$

The same with $v=\alpha w$.

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