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!

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

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  • $\begingroup$ 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? $\endgroup$ – thenark May 4 '19 at 13:01
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    $\begingroup$ 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 $B\Rightarrow A$. $\endgroup$ – tia May 4 '19 at 13:05
  • $\begingroup$ 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 $||u+v||^{2}=(u+v)\cdot(u+v)$? $\endgroup$ – thenark May 4 '19 at 23:07
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    $\begingroup$ I edited the answer to include more details. Let me know if this helps. $\endgroup$ – tia May 5 '19 at 1:42
  • $\begingroup$ Thank you so much, @tia! $\endgroup$ – thenark May 5 '19 at 11:53

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

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  • $\begingroup$ 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? $\endgroup$ – thenark May 4 '19 at 12:50
  • $\begingroup$ If two vectors $u$ and $v$ have the same direction then $\cos(u,v)=\pm1$. $\endgroup$ – DINEDINE May 4 '19 at 12:52
  • $\begingroup$ Thanks for this. I've apparently been looking at arccos in my calculator the past hour, which is why pi/4 keeps showing up. $\endgroup$ – thenark May 4 '19 at 12:56
  • $\begingroup$ 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. $\endgroup$ – tia May 4 '19 at 13:09

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