1
$\begingroup$

I'm trying to show the following inequality, which expresses the fact that the magnitude of the sum of some vectors, is less than the sum of the individual magnitudes: $$\left\lVert\sum_i \vec{a_i}\right\rVert \le \sum_i \left\lVert \vec{a_i}\right\rVert$$ I found this on a page with Cauchy Schwarz problems, but stated without proof.

I've also be considered if it can be solved by induction, but I haven't got an idea how.

Improve this question
$\endgroup$
  • $\begingroup$ This sum has finite number of terms? $\endgroup$ – user73454 May 10 '14 at 15:27
  • $\begingroup$ Yes, sorry for not making that clear. $\endgroup$ – Thomas Ahle May 10 '14 at 16:06
3
$\begingroup$

Triangle Inequality: $\|v+w\|\leq \|v\|+\|w\|$. Thus, $$ \|\sum\limits_{i=1}^n v_i\|=\|\sum\limits_{i=1}^{n-1} v_i+v_n\|\leq\| \sum\limits_{i=1}^{n-1} v_i\|+\|v_n\|, $$ then applies induction, i.e., $\| \sum\limits_{i=1}^{n-1} v_i\|\leq \sum\limits_{i=1}^{n-1} \|v_i\|$.

Improve this answer
$\endgroup$
  • $\begingroup$ Doh, you are right. I guess I was just specifically looking for a Cauchy Schwarz proof. $\endgroup$ – Thomas Ahle May 10 '14 at 15:49
  • 1
    $\begingroup$ @ThomasAhle, Let $t=sign(\langle v, w\rangle)$, then $$\| v+tw\|^2=\| v\|^2+2t\langle v, w\rangle +\|w\|^2,$$ but $\|v+tw\|\let \|v\|+\|w\|$, then $$\|v+tw\|^2\leq\|v\|^2+2\|v\|\|w\| +\|w\|^2. $$ Thus $$t\langle v,w\rangle\leq \|v\|\|w\|$$ i.e., $$|\langle v,w\rangle|\leq \| v\|\|w\|.$$ $\endgroup$ – user73454 May 10 '14 at 16:30
  • $\begingroup$ @ThomasAhle, there is problem on to compile, but let $t=sign(\langle v,w\rangle)$, then $$|v+tw|^2=|v|^2+2t\langle v,w\rangle+|w|^2$$ but $|v+tw|\leq |v|+|t||w|\leq |v|+|w|$, then $$|v + tw|^2\leq |v|^2+2|v||w|+|w|^2.$$ Thus $$t\langle v,w\rangle=|\langle v,w\rangle|\leq |v||w|.$$ $\endgroup$ – user73454 Dec 29 '14 at 18:20

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.