Q:

Prove: $|\langle u,v \rangle|= \|u\|\cdot \|v\|\iff u=\alpha v$

Let $V$ be an inner product space s.t $u,v\in V$ and $u,v\neq 0$
Prove:  $$\langle u,v \rangle = ||u||\cdot ||v||\iff u=\alpha v$$ s.t $\alpha\in \mathbb{C}$

$\Leftarrow:$
$$|\alpha| \langle  v,v \rangle=|\alpha\langle  v,v \rangle|=|\langle  \alpha v,v \rangle|=|\langle u,v \rangle|=|\langle u,v \rangle| \leq |\|u\|\cdot\|v\| |=| \|\alpha v\|\cdot\|v\| |=|\alpha |||v||^2=|\alpha| \langle  v,v \rangle$$
I am left with $|\alpha|$
How should I approach the proof of $\Rightarrow$

A:

The statement is incorrect.

You can consider the orthogonal projection of $u$ on the subspace generated by $v$: then $u=\alpha v+w$, where $\langle w,v\rangle=0$.
Now, assuming linearity in the first variable,
$$
\langle u,v\rangle=\langle \alpha v+w,v\rangle=\alpha\langle v,v\rangle
$$
Also
$$
\|u\|^2=\langle \alpha v+w,\alpha v+w\rangle=|\alpha|^2\langle v,v\rangle+\langle w,w\rangle
$$
Assuming $\|u\|\cdot\|v\|=\langle u,v\rangle$, we obtain
$$
\alpha^2\langle v,v\rangle^2=
(|\alpha|^2\langle v,v\rangle+\langle w,w\rangle)\langle v,v\rangle
$$
that simplifies to
$$
(\alpha^2-|\alpha|^2)\|v\|=\|w\|
$$
In particular $\alpha^2$ must be real and nonnegative, which forces $\alpha$ to be real, so $\alpha^2=|\alpha|^2$ and therefore $\|w\|=0$.
It would be similar with linearity in the second variable.

For the $\Leftarrow$ direction, if $u=\alpha v$, then
$$
\langle u,v\rangle=\alpha\|v\|
$$
and $\|u\|\,\|v\|=|\alpha|\,\|v\|$, so we get the equality only if $\alpha$ is real and nonnegative.

A correct statement would be:

Let $V$ be a real inner product space, $u,v\in V$ be nonzero vectors. Then $\langle u,v\rangle=\|u\|\,\|v\|$ if and only if $u=\alpha v$ for a positive scalar $\alpha$.

