4
$\begingroup$

I'm currently taking a university course in Linear Algebra and Matrix Theory. A recent problem set included a question that asked,

What can you say about two nonzero vectors $\vec{\alpha}$ and $\vec{\beta}$ that satisfy the equation: $$\|\vec{\alpha}+\vec{\beta}\| \ = \ \|\vec{\alpha}\| + \|\vec{\beta}\| \ $$ $$\vec{\alpha},\vec{\beta} \in \mathbb{R}^n$$

I am attempting to solve this by finding a solution from this equation derived from the law of cosines: $$\|\vec{\alpha}+\vec{\beta}\|^2 \ = \ \|\vec{\alpha}\|^2 + \|\vec{\beta}\|^2 - \ 2\|\vec{\alpha}\| \|\vec{\beta}\|\cos(\pi-\theta)$$ ...so far I have been unable to find a valid solution and am tempted to assert that there exists no $\vec{\alpha}$ and $\vec{\beta}$ for which that equation is true.

Is there any case in which the magnitude of the sum of two vectors equals the sum of the magnitudes?

$\endgroup$
0
7
$\begingroup$

Working from your point of view you want, for nonzero $\vec\alpha$ and $\vec\beta$, $$ (\|\vec\alpha\|+\|\vec\beta\|)^2=\|\vec\alpha+\vec\beta\|^2=\|\vec\alpha\|^2+\|\vec\beta\|^2-2\|\vec\alpha\|\, \|\vec\beta\|\,\cos(\pi-\theta). $$ So you need $\cos(\pi-\theta)=-1$, which is exactly $\theta=0$. So the two vectors are colinear.

$\endgroup$
3
  • $\begingroup$ ... if their scalar values are non zero. If they're zero they could be pointing any old which a way, which is to say, not pointing at all. $\endgroup$ Sep 25 '16 at 4:10
  • $\begingroup$ Yes, good point. $\endgroup$ Sep 25 '16 at 4:55
  • $\begingroup$ This makes total sense, it never occurred to me that you could simply "reverse foil" the expression when the cosine expression equals negative one. Thank you! $\endgroup$ Sep 25 '16 at 23:13
4
$\begingroup$

Yes. This occurs exactly when one vector is a nonnegative scalar multiple of the other.

$\endgroup$
0
2
$\begingroup$

In $\mathbb{R}^n$, we have that your equality implies $$\langle a+b,a+b \rangle=(\Vert a\Vert+\Vert b \Vert)^2 ,$$ which, using the bilinearity of the inner product, yields $$\Vert a \Vert^2+\Vert b \Vert^2 +2\langle a,b \rangle=\Vert a\Vert^2+\Vert b \Vert ^2 +2\Vert a \Vert \Vert b \Vert$$ $$\implies \langle a,b \rangle= \Vert a \Vert \Vert b \Vert.$$ And equality on Cauchy-Schwarz only holds if both vectors are linearly dependent. (Note that since there is no modulus on the inner product on the left side, not only they must be linearly dependent, but also differ by a positive scaling).

As a sidenote, the technique above holds for any inner product space. One might be tempted to extend it to any normed space, but the result isn't true. As an example, one can take $L^1([0,1])$ and $a=I_{[0,1/2]}$, $b=I_{[1/2,1]}$, where $I_A$ is the indicator function on $A$.

$\endgroup$
1
$\begingroup$

Recall that $\|\vec x\|^2 = \vec x \cdot \vec x$.

If one vector is zero then the equation is trivially true, so suppose they are nonzero. If you square both sides you get $$ \| \vec \alpha \|^2 + 2(\vec \alpha \cdot \vec \beta) + \| \vec \beta \|^2 = \| \vec \alpha \|^2 + 2\|\vec \alpha\| \|\vec \beta\| + \| \vec \beta \|^2$$ so that $\vec \alpha \cdot \vec \beta = \|\vec \alpha\| \|\vec \beta\|$. This can only happen if the two vectors point in the same direction. (To see this, look at the Cauchy-Schwarz inequality says, or compute the angle $$\theta = \arccos \frac{\vec \alpha \cdot \vec \beta}{\|\vec \alpha\| \|\vec \beta\|}$$ between the two vectors.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.