I'm trying to show that $\|u\cdot v\|= \|u\| \, \|v\| \iff u = \alpha v, \quad \alpha\in \mathbb{R}$.
$(\Leftarrow)\;$ Note that $\|\alpha v\cdot v\| = |\alpha| \, \| v\cdot v\| = |\alpha| \, \| (\sum_{k=1}^n v_k^2)\| = |\alpha| \, \sum_{k=1}^n v_k^2$. Also, $\|v\| = \sqrt{\sum_{k=1}^n v_k^2}$, so $\|u\| = \|\alpha v\| = |\alpha| \, \|v\| = |\alpha| \sqrt{\sum_{k=1}^n v_k^2}$.
Therefore, $\|u\cdot v\| \, = |\alpha| \sqrt{\sum_{k=1}^n v_k^2} \, \sqrt{\sum_{k=1}^n v_k^2} = |\alpha| \sum_{k=1}^n v_k^2 = |\alpha| \, \|v \cdot v\| = \|\alpha v \cdot v\|$.
$(\Rightarrow)\;$ We see that $\|u \cdot v\| = \bigl\| \sum_{k=1}^n {u_k v_k} \bigr\| = \bigl| \sum_{k=1}^n {u_k v_k} \bigr|, \qquad \|u\| = \sqrt{\sum_{k=1}^n u_k^2}, \qquad \|v\| = \sqrt{\sum_{k=1}^n v_k^2}$.
Then $\|u\| \, \|v\| = \sqrt{\sum_{k=1}^n u_k^2} \, \sqrt{\sum_{k=1}^n v_k^2} = \sqrt{\sum_{k=1}^n u_k^2 \, \sum_{k=1}^n v_k^2} = \sqrt{\sum_{k=1}^n \sum_{k=1}^n u_k^2 \, v_k^2}$.
I got stuck at this point trying to show that $\sqrt{\sum_{k=1}^n u_k^2} \, \sqrt{\sum_{k=1}^n v_k^2} = \bigl| \sum_{k=1}^n {u_k v_k} \bigr|$.
Can someone lend me a hand, please?