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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!!

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  • $\begingroup$ Were you trying to find a counterexample $\endgroup$
    – Amr
    Dec 25, 2013 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, 2013 at 17:35

3 Answers 3

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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.

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  • $\begingroup$ but you show that they are linearly independent, I needed to show that they are linearly dependant $\endgroup$
    – gbox
    Dec 25, 2013 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, 2013 at 13:42
  • $\begingroup$ @miracle173: The answer has now been edited, so you should be able to undownvote, $\endgroup$ Dec 27, 2013 at 14:04
  • $\begingroup$ @Henning Makholm: Thanks, done $\endgroup$
    – miracle173
    Dec 29, 2013 at 1:47
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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 $0$ ).....

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  • $\begingroup$ but to prove p$\rightarrow$q I need to contradict $(p\wedge\neg q)$ $\endgroup$
    – gbox
    Dec 25, 2013 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, 2013 at 21:16
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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 $\{u_1+u_2, u_2+u_3, u_1+u_3\}$?

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