Proving linear dependency for two vector groups The question:
Let V be a vector space over $\mathbb{R}$.
Let $S = \{v,u,w\}$ be a group of 3 vectors in V.
Let T be defined as $T = \{v, 
v + u,
v + u + 2w
\}$.
Prove that if S is linearly dependent, then also T is linearly dependent.

The proof:
Let's assume that S is linearly dependent, and assume to the contrary that T is linearly independent.  
Thus, for $\lambda_1,\lambda_2,\lambda_3 \in \mathbb{R}$, if the following exists:
$$\lambda_1v + \lambda_2(v + u) + \lambda_3 (v + u + 2w) = 0$$
Then, necessarily $\lambda_1 = \lambda_2 = \lambda_3 = 0$.
Rearranging the equation:
$$(\lambda_1 + \lambda_2 + \lambda_3)v + (\lambda_2 + \lambda_3)u + 2\lambda_3w = 0$$
But it is given that S is linearly dependent, therefore the above equation is suppose to be true also for $\lambda_1,\lambda_2,\lambda_3$ which aren't all zero.
This is in contradiction that T is linearly dependent, therefore T must be linearly independent.
Q.E.D.

Is the proof correct?
If not, why?
Could you forumlate a proof in a similar method which is correct?
 A: This would be a valid way of doing the problem:
We want to show that if $T$ is linearly independent, then so is $S$ (thus proving the desired statement by contrapositive).  Suppose $T$ is linearly independent.  Let $c_i$ be such that
$$
c_1 v + c_2 u + c_3 w = 0
$$
We would like to show that all $c_i$ are zero, thus proving that $S$ is linearly independent.  We note that
$$
c_1 v + c_2 u + c_3 w = \\
(c_1 - c_2)v
+(c_2 - c_3/2)(v+u)
+(c_3/2)(v + u + 2w) = 0
$$
Because $T$ is linearly independent, we may conclude that
$$
c_1 - c_2 = 0\\
c_2 - c_3/2 = 0\\
c_3/2 = 0
$$
From the above, we may conclude that $c_1 = c_2 = c_3 = 0$.  Thus, $S$ is linearly independent, as desired.
A: You don't really have a contradiction, until you give a way for providing a set of coefficients $\lambda_1,\lambda_2,\lambda_3$ that are not all zero.
There's a simpler way: the span of $T=\{v,v+u,v+u+2w\}$ is clearly contained in the span of $S=\{u,v,w\}$, because
$$
\lambda_1v + \lambda_2(v + u) + \lambda_3 (v + u + 2w) =
(\lambda_1 + \lambda_2 + \lambda_3)v + (\lambda_2 + \lambda_3)u + 2\lambda_3w
$$
as you computed.
Since the dimension of the span of $S$ is at most $2$, because $S$ is linearly dependent, the span of $T$ has dimension at most $2$. Therefore $T$ is linearly dependent.

The objection that $u,v,w$ are meant to be distinct is invalid. For instance, if $w=v$, we'd have $T=\{v,v+u,u+3v\}$, but again this would be linearly dependent. The only way it couldn't be is that the three expressions denote the same non zero vector, because the span of $S$ has dimension at most $1$. Let's see: $v=v+u$ entails $u=0$ and $v=u+3v$ entails $v=0$. So certainly $T$ is linearly dependent.
Similarly, if $w=u$ we have $T=\{v,v+u,v+3u\}$ and the same reasoning holds.
If $u=v$, we have $T=\{u,2u,2u+2w\}$ and so $u=2u$ means $u=0$.
But there's really no need to do this analysis.
