Answer
False.
Work Step by Step
Suppose $\vec{v}_{1}=(0,0,0,0)$ and $\vec{v}_{2}=(1,1,1,1)$. Then $\vec{v}_{2}$ is not a scalar multiple of $\vec{v}_{1}$, since $k\cdot\vec{0}=\vec{0}$ for all scalars $k$, but the set containing the vectors is nevertheless linearly dependent, since the zero vector is in the set.
