If a set of vectors is dependent, there is a non-trivial combination of some of them that equals 0: $a_1v_1+\dots+a_n v_n=0,$ where not all the scalars $a_i$ are $0$. Linearity of $T$ should give you at once that the $Tv_i$ are also linearly dependent (as witnessed by the same $a_i$).
Linear independence, on the other hand, does not need to be preserved. For example, consider the linear transformation that maps all the vectors to 0.
Now, under some additional conditions, a linear transformation may preserve independence. For example, suppose that $T$ is injective (i.e., the only solution to $Tv=0$ is $v=0$). Then $T$ preserves linear independence: Suppose that $Tv_1,\dots,Tv_n$ are dependent, so there are scalars $b_1,\dots,b_n$, not all zero, such that $b_1Tv_1+\dots+b_nTv_n=0.$ By linearity, you get $T(b_1v_1+\dots+b_nv_n)=0$ and, if $T$ is injective, then in fact $b_1v_1+\dots+b_nv_n=0$, so the $v_i$ are dependent. (Or, if you prefer: If the $v_i$ are independent, and we have $b_1Tv_1+\dots+b_nTv_n=0$, then as above we get $b_1v_1+\dots+b_nv_n=0$, and independence gives us that $b_1=\dots=b_n=0$, so the $Tv_i$ are independent.)
Injectivity of $T$ is the only way to ensure that any independent set is mapped to an independent set. However, even if $T$ is not injective, there may be some independent sets whose independence is preserved. For example, consider the linear transformation $T:{\mathbb R}^3\to{\mathbb R}^2$ given by $T(x,y,z)=(x,y)$.
Then $T$ maps both $(1,0,0)$ and $(1,0,1)$ to $(1,0)$. The original vectors are independent, the resulting vectors obviously are not (they coincide!). But, on the other hand, $T$ preserves the linear independence of any two (independent) vectors whose last coordinate is 0.