Show that one gets equality in the Schwarz inequality if and only if $\overrightarrow v$,$\overrightarrow w$ are linearly dependent.
(I am supposing they want me to prove it in an inner product space we call V)
The $\Leftarrow$ part of the proof was pretty straightforward but I don't know how to go about the $\Rightarrow$ part because linear dependence requires us to show $\overrightarrow v$=$\lambda \overrightarrow w$ for some $\lambda \in \mathbb{R}$ but the Cauchy-Schwarz inequality only involves norms\lengths of vectors and not relations between the vectors themselves. I would appreciate tips on starting points.
Best Answer
Assume $\overrightarrow v\neq0$(if $\overrightarrow v=0$ the proof is trivial), and take any norm of $V$. Fix an arbitrary $\lambda$ in the scalar field ($\mathbb{R}$in this case) and note that $$0\leq \|\overrightarrow u-\lambda \overrightarrow v \|^2=\langle \overrightarrow u-\lambda \overrightarrow v ,\overrightarrow u-\lambda \overrightarrow v \rangle$$ Is the first step in deriving the C.S inequality. Thus equality holds iff $$\|\overrightarrow u-\lambda \overrightarrow v \|=0$$ From our properties of norms we know then that $$\overrightarrow u-\lambda \overrightarrow v=0 $$ and we have shown linear dependence.