Prove that $\|u + v\| =\|u\| + \|v\|$ if and only if $u$ and $v$ have the same direction.

Question

I'd like some help with a proving problem I have in my linear algebra textbook. My background: I'm a business student taking a linear algebra class and I have a really hard time with proving problems in general. Any nudge in the right direction would be greatly appreciated!

Question: Prove that $\|u + v\| = \|u\| + \|v\|$ if and only if $u$ and $v$ have the same direction.

Here's what I have so far:

Let $u$ and $v$ be two vectors who share the same direction.
$u = cv$
$u = \frac {v}{\|v|\|}$
$u+v = \frac {v}{\|v\|} + v$
$\|u+v\| = \|\frac {v}{\|v\|} + v\|$

At this point, I don't really understand what I'm doing anymore.

Another thing I tried doing:
$\|u + v\|^{2} = (u+v)\cdot(u+v)$
$\|u + v\|^{2} = u\cdot(u+v) + v\cdot(u+v)$
$\|u + v\|^{2} = \|u\|^{2} + 2(u\cdot v) + \|v\|^{2}$

After this I'm kind of stuck. How do I prove that $(u\cdot v) = 0$? I tried using the fact that the $\theta = 0$ since the two vectors are parallel(?), which I then plug into: $cos\theta = \frac{(u\cdot v)}{\|u|\|\|v\|}$ but that doesn't really lead me anywhere.

Edit: Just realised that $(u\cdot v) = 0$ will only be the case if $u$ and $v$ are orthogonal–which is definitely not the case. I also just realised that the proof I've attempted to do is similar to the proof for the Triangle Inequality. I also wouldn't know how to get rid of the squares.

Would I need to use contradiction since I need to prove an if and only if statement?

Thank you!

Answer 1

There are many ways to do this, and squaring is one of them. Note that when $u,v$ have the same direction, then $$u\cdot v = \left \| u \right \|\left \| v \right \| cos(0) = \left \| u \right \|\left \| v \right \|.$$ So, $$||u + v||^{2} = ||u||^{2} + 2(u\cdot v) + \|v\|^{2} = \|u\|^{2} + 2\left \| u \right \|\left \| v \right \| + \|v\|^{2} = (\|u\| + \|v\|)^2.$$

Taking the square roots at both sides, you get that $$\|u+v\| = \|u\|+\|v\|.$$

Conversely, if $\|u+v\| = \|u\|+\|v\|$, then squaring you get that $$||u||^{2} + 2(u\cdot v) + \|v\|^{2} = \|u\|^{2} + 2\left \| u \right \|\left \| v \right \| + \|v\|^{2} \Rightarrow \\2\| u \|\left \| v \right \| cos(\theta) = 2\| u\| \| v\| \Rightarrow cos(\theta) = 1 \Rightarrow \theta = 0.$$

Answer 2

$$||u+v||^2=(||u||+||v||)^2\iff 2uv=2||u||.||v||\iff\cos(u,v)=\pm1$$