Prove that if two vectors are linearly dependent, their sums are also linearly dependent If {$u,v$} is linearly dependent, prove {$u,u+v$} is also linearly dependent. I have so far that if $du+c(u+v)=du+cu+cv=(d+c)u+cv=0$ then at least $d+c\not =0$ or $c\not=0$ since {$u,v$} is linearly dependent. But then I seem to get caught in a fairly large amount of cases to consider. I'm wondering if there is a more efficient way to prove this or a theorem perhaps?
 A: If $\{u,v\}$ are linearly dependent then $v=cu$ for some $c$ (unless $u=\vec{0}$ and $v\ne \vec{0}$). So
$$u+v=u+cu=(1+c)u,$$ from where you have that $u,u+v$ are linearly dependent. 
If $u=\vec{0}$ then $u+v=v$ and $u=\vec{0}$ are linearly dependent. Note that $u=0(u+v).$ 
A: Proof by contrapositive
Suppose $\{u,u+v\}$ is linearly independent. Take scalars $a$ and $b$ such that $au+bv=0$; then $(a-b)u+b(u+v)=0$. Since $\{u,u+v\}$ is linearly independent, we obtain $a-b=0$ and $b=0$, so $a=0$. Therefore $\{u,v\}$ is linearly independent.
Direct proof
Let $a$ and $b$ be scalars, not both zero, such that $au+bv=0$; then $(a-b)u+b(u+v)=0$. If $b=0$, then $a\ne0$ and so $a-b\ne0$ as well.
More general statement
(Note: we're talking about lists of vectors, rather than sets, as usual when linear dependence and independence is dealt with.)
Suppose the vectors $u_1,\dots,u_n$ are linearly dependent and that $w_1,\dots,w_n$ (same number) are linear combination of those vectors, Then also $w_1,\dots,w_n$ are linearly dependent. Indeed, $\dim\operatorname{span}\{u_1,\dots,u_n\}<n$, so
$$
\dim\operatorname{span}\{w_1,\dots,w_n\}\le
\dim\operatorname{span}\{u_1,\dots,u_n\}< n
$$
