# 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. – Git Gud Nov 11 '14 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 :) – user860374 Nov 11 '14 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 $v=\alpha w$.