Equality in the Cauchy-Schwarz inequality implies $\overrightarrow v$,$\overrightarrow w$ linearly dependent

Question

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.

Answer 1

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.

Answer 2

From the proof of C-S. With $0\ne x=(x_1,..,x_n)$ and $0\ne y=(y_1,..,y_n).$ $$\|x\|^2\cdot \|y\|^2-(x\cdot y)^2=\sum_{1\leq i<j\leq n}(x_i y_j-x_j y_i)^2.$$ Now $x_i y_j=x_j y_i$ whenever $1\leq i<j\leq n$ iff there exists $k\ne 0$ such that $x_i=k y_i$ for $i=1,..,n.$ Use induction on $n$ to prove this.

The case $x=0\lor y=0$ is trivial.