1
$\begingroup$

I had to prove the following notation: if $u1,u2,u3$ are linear depended then $u1+u2,u2+u3,u3+u1$ are also linear depended.

I tried to contradict saying that I assume that P is right but Q is wrong as for $u1,u2,u3$ are linear depended and $u1+u2,u2+u3,u3+u1$ are not linear depended.

I choose $u1 =(0,0,1), u2= (1,1,0), u3=(2,2,1)$ and show that both $u1,u2,u3$ and $u1+u2,u2+u3,u3+u1$ are linear depended.

  1. Is my proof is right?
  2. what is the reason that the notation is right?

Thanks a lot!!

$\endgroup$
  • $\begingroup$ Were you trying to find a counterexample $\endgroup$ – Amr Dec 25 '13 at 17:32
  • $\begingroup$ But, unless your looking for a counterexample, your argument has to hold for any triple of linearly-independent vectors. $\endgroup$ – user99680 Dec 25 '13 at 17:35
5
$\begingroup$

Your work is wrong since in a proof an example isn't sufficient.

Take a linear combination of these vectors: $$\alpha(u_1+u_2)+\beta(u_2+u_3)+\gamma(u_3+u_1)=0 $$ hence $$(\alpha+\gamma)u_1+(\alpha+\beta)u_2+(\beta+\gamma)u_3=0$$ but since $u_1,u_2,u_3$ are linearly independant then $$\alpha+\gamma=\alpha+\beta=\beta+\gamma=0$$ and hence it's easy to show that $$\alpha=\beta=\gamma=0$$ and we deduce.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ but you show that they are linearly independent, I needed to show that they are linearly dependant $\endgroup$ – gbox Dec 25 '13 at 18:17
  • $\begingroup$ i noticed that i have downvoted this by accident about an hour ago. It is not possible to undo this downvote now. sorry. $\endgroup$ – miracle173 Dec 27 '13 at 13:42
  • $\begingroup$ @miracle173: The answer has now been edited, so you should be able to undownvote, $\endgroup$ – hmakholm left over Monica Dec 27 '13 at 14:04
  • $\begingroup$ @Henning Makholm: Thanks, done $\endgroup$ – miracle173 Dec 29 '13 at 1:47
2
$\begingroup$

Let $u_1,u_2,u_3$ be linearly-indepent vectors, and assume {$u_1+u_2, u_2+u_3 , u_3+u_1$} are linearly-dependent. Then there is a triple $(c_1,c_2,c_3)$ of Reals, not all $0$ , with: $$c_1(u_1+u_2)+c_2(u_2+u_3)+c_3(u_1+u_3)=(c_1+c_3)u_1+ (c_1+c_2)u_2+ (c_2+c_3)u_3=0 $$ (and note that, for the sake of completeness, that if $c_1,c_2,c_3$ are not all $0$, the triple $(c_1+c_3, c_1+c_2, c_2+c_3)$ cannot all be $0$ ).....

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ but to prove p$\rightarrow$q I need to contradict $(p\wedge\neg q)$ $\endgroup$ – gbox Dec 25 '13 at 18:32
  • $\begingroup$ But , the set S:={$u_1+u_2, u_2+u_3,u_1+u_3$} is actually independent, so you cannot prove they are dependent. Isn't that what is proved here? You assume $p$: the vectors $u_1.u-2, u_3$ are independent, you assume $q$: the set S is dependent, and you arrive at a contradiction. $\endgroup$ – user99680 Dec 25 '13 at 21:16
0
$\begingroup$

Hint:

Let $S = \{u_1, u_2, u_3\}$ be the set containing the given vectors. Now, consider $\rm{span}\{S\}$. Since $S$ is linearly dependent, it follows that the basis of $\rm{span}\{S\}$ is a proper subset of $S$ (that is, a subset of $S$ with fewer elements).

Now, what are the possible dimensions of $\rm{span}\{S\}$? How do they compare to the size of the set $\{u_1+u_2, u_2+u_3, u_1+u_3\}$?

| cite | improve this answer | |
$\endgroup$

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.