2
$\begingroup$

I've solved the following problem: "Let $u,v \in \mathbb{R}^n$ with $u \neq 0$ be such that $|u+v|=|u|+|v|$ (euclidean norm), show that there's $\alpha \in \mathbb{R}$ with $\alpha \geq 0$ such that $v = \alpha u$".

My approach was the following: By definition of the euclidean norm, we have that $$|u+v|^2=\left\langle u+v,u+v\right\rangle=|u|^2+2\left\langle u,v\right\rangle+|v|^2,$$ but by hypothesis $|u+v|^2=(|u|+|v|)^2=|u|^2+2|u||v|+|v|^2$ so that equating those two we must have $\left\langle u,v\right\rangle = |u||v|$ and so $|\left\langle u,v\right\rangle| =|u||v|$ so that by the Cauchy-Schwarz inequality there must be $\alpha \in \mathbb{R}$ such that $v=\alpha u$.

Since I've not shown that $\alpha \geq 0$ in this proof, I thought it was wrong, but the answer in the book gives the same proof and says that all of this implies $\alpha \geq 0$. I've gone again through the proof of Cauchy-Schwarz Inequality and it doesn't seem to be there the reason for this. Indeed, if $x,y \in \mathbb{R}^2$ are simply $x=(1,1)$ and $y=(-2,-2)$ then $y=-2x$ and indeed $|\left\langle x,y\right\rangle| = |x||y|$, so by this example we see that Cauchy-Schwarz Inequality doesn't imply that the scalar that multiply one vector to get the other must be positive.

So, how this $\alpha$ is shown to be greater or equal to zero?

Thanks very much in advance!

EDIT: I think I've found the answer. I've shown that $\left\langle u,v\right\rangle =|u||v|$ and thus $|\left\langle u,v\right\rangle| = |u||v|$ by Cauchy-Schwarz Inequality we must have $v=\alpha u$, but since $\alpha = \left\langle u,v\right\rangle /|u|^2$, since the inner product itself is a positive number (because it's a product of two norms), then $\alpha$ must be positive. Is this right?

$\endgroup$
1
  • 1
    $\begingroup$ That's absolutely right. $\endgroup$ Aug 10 '13 at 18:06
3
$\begingroup$

You are correct: $\alpha \ge 0$ should be shown independently.

And your edit is also correct.

$\endgroup$

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.