Sum of Linearly Dependent Vectors Is it possible that vectors $v_1, v_2, v_3$ are linearly dependent, but the 
vectors $w_1=v_1+v_2$, $w_2=v_1+v_3$, $w_3=v_2+v_3$ are linearly independent? I 
believe the answer is no, this is not possible, but I am struggling to formulate 
the proof.

No, they must be dependent. Note that you have
$$\begin{align}
v_1 &= \frac{1}{2}(w_1+w_2-w_3) \\
v_2 &= \frac{1}{2}(w_1-w_2-w_3) \\
v_3 &= \frac{1}{2}(-w_1+w_2-w_3) 
\end{align}$$
Now, by assumption the $v_i$'s linearly dependent, so that there exist $(a,b,c)\neq(0,0,0)$ such that $0=av_1 + bv_2+c v_3$. This gives, if I didn't mess up,
$$
0 = (a+b-c) w_1 + (a-b+c) w_2+ (-a+b+c) w_3
$$
which will prove the $w_i$'s are linearly dependent, once we have shown that at least one of the 3 resulting coefficients is non-zero. But indeed, this must hold: otherwise, if they all are equal to zero, solving the system
$$\begin{align}
0 &= a+b-c \\
0 &= a-b+c \\
0 &= -a+b+c
\end{align}$$
shows there is only one solution, $(a,b,c)=(0,0,0)$: which was excluded by assumption.
