Proving that $|\langle x,y\rangle|=\|x\|\|y\|$ iff $x,y$ are linearly dependent

Question

Show: For any vectors $x$ and $y$ in an inner product space $V$ over $\mathbb{F}$, we have $$\left|\langle x,y\rangle\right|=\|x\|\|y\| \iff x,y \ are \ dependent$$

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($\leftarrow$) If $x$ and $y$ are dependent, then $x=ky$, $k\in \mathbb{F}$. So, $$|\langle x,y\rangle|=|\langle kx,y\rangle|=|k|\|y\|^2=|k|\|y\|\|y\|=\|ky\|\|y\|=\|x\|\|y\| $$

How, do I prove ($\rightarrow$)?

Answer 1

Assume that $y \neq 0$, otherwise there's not much to show. If $y \neq 0$, let $x_{//} = \frac{\langle x, y\rangle}{\langle y, y\rangle} y$ and $x_{\perp} = x - x_{//}$. Show that $\langle x, y\rangle = \langle x_{//}, y \rangle$ and $\| x_{//}\| \leq \| x \|$ with equality if and only if $x = x_{//}$. Therefore, $| \langle x, y \rangle| = |\langle x_{//}, y \rangle| \leq \|x_{//}\| \|y\|$, but by assumption $|\langle x, y\rangle| = \|x\|\|y\|$ which will imply that $\|x\|=\|x_{//}\|$ and hence $x = x_{//}$. Now, you have that $x_{//}$ is linearly dependent on $y$.

Answer 2

In case $x\neq0$, show that the length of the vector $y-\dfrac{\langle x,y\rangle}{\|x\|^2}x$ equals zero.

Answer 3

Note that the assertion holds trivially when $x=0$ or $y=0$.

Let $c\in \mathbb R$ such that $\|y\|=c\|x\|$. Then $$ \|cx-y\|^2=\langle cx-y,cx-y\rangle=\|cx\|^2+\|y\|^2-2c \langle x,y\rangle\\=\|cx\|^2+\|y\|^2-2c\|x\|\,\|y\|=\|cx\|^2+\|y\|^2-2\|cx\|\,\|y\|\\=(\|cx\|-\|y\|)^2=(c\|x\|-\|y\|)^2=0. $$ So $\|cx-y\|=0$, and this means that $y=cx$.