text: # Cauchy-Schwarz inequality with zero angle?
  
  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?
  
  • 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. Nov 11, 2014 at 16:37
  • @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 :) Nov 11, 2014 at 16:39
  
  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 $$1=\frac{\langle u,v\rangle}{\|u\| \|v\|}\iff \langle u,v\rangle = \|u\| \|v\|$$
  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 $v=\alpha w$.