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

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!

• Please note that there are norms for which the claim isn't true; the Manhattan-norm would be an example. (However, it's true for any norm which derives from an inner product.) – Michael Hoppe May 4 '19 at 13:43

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 $$||u||^{2} + 2(u\cdot v) + \|v\|^{2} = \|u\|^{2} + 2\left \| u \right \|\left \| v \right \| + \|v\|^{2} \Rightarrow \\2\| u \|\left \| v \right \| cos(\theta) = 2\| u\| \| v\| \Rightarrow cos(\theta) = 1 \Rightarrow \theta = 0.$$

• Thanks for the clear and concise explanation! May I ask if you need to prove "two sides" if proving an if and only if statement? – thenark May 4 '19 at 13:01
• Yes. A if and only if B is another way of saying that A and B are equivalent. To prove this, you have to show that $A\Rightarrow B$ and $B\Rightarrow A$. – tia May 4 '19 at 13:05
• Hi, thanks again for the answer! Just one last clarification. I tried expanding the second portion of the proof and ended up with: $||u||^{2}+2||u||||v||+||v||^{2}$. How do I jump from here to proving that such an equation only holds true if $\theta = 0$? Can I just say that or do I need to go back all the way to $||u+v||^{2}=(u+v)\cdot(u+v)$? – thenark May 4 '19 at 23:07
• I edited the answer to include more details. Let me know if this helps. – tia May 5 '19 at 1:42
• Thank you so much, @tia! – thenark May 5 '19 at 11:53

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

• Sorry, I don't really understand how this goes. I also just realised that it only becomes 0 if u and v are orthogonal which is not the case. Also, if the two vectors share the same angle, then isn't cos(0) = pi/4? – thenark May 4 '19 at 12:50
• If two vectors $u$ and $v$ have the same direction then $\cos(u,v)=\pm1$. – DINEDINE May 4 '19 at 12:52
• Thanks for this. I've apparently been looking at arccos in my calculator the past hour, which is why pi/4 keeps showing up. – thenark May 4 '19 at 12:56
• If $cos(u,v) = -1$ then $u\cdot v = -\|u\|\|v\|$ and $\|u+v\| = |\|u\|-\|v\||$. In this case, $u$ and $v$ are colinear, but they have opposite directions. – tia May 4 '19 at 13:09