Linear Dependence Of A Sum 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.


*

*Is my proof is right?

*what is the reason that the notation is right?


Thanks a lot!!
 A: 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.
A: 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 $0$ ).....
A: 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 $\{u_1+u_2, u_2+u_3, u_1+u_3\}$?
