Linear Transformation of Dependent set. Let $V$ and $W$ be vector spaces and let $T: V\to W$ be a linear transformation. Let $\{v_1, v_2,\ldots, v_p\}$ be a linearly dependent set of vectors in $V$. Show that $\{Tv_1, Tv_2,\ldots, Tv_p\}$ is also linearly dependent.
 A: Prove that if $T(v_1),\ldots,T(v_p)$ are linearly independent, then so are $v_1,\ldots,v_p$. 
A: It's easier to prove the contrapositive statement.  Suppose $T(v_i),..., T(v_p)$ are linearly independent and let $\sum_{n=1}^{p}a_iv_i=0$ . Now applying $T$ we get $T(\sum_{n=1}^{p}a_iv_i)=T(0)$. But since $T$ is linear it follows that $\sum_{n=1}^{p}a_iT(v_i)=0$. But since $T(v_i),..., T(v_p)$ are linearly independent, $a_i=0$ for all $i \in \{1,2,....,p\}$. Thus $v_i,......,v_p$ are linearly independent.
A: If $\{v_1, v_2,\ldots, v_p\}$ is linearly dependent, there exists $a_1, \cdots, a_p \in \mathbb{R}$, such that
$$a_1v_1+\cdots+a_pv_p =0 $$
where $a_i \neq 0$, for at least one $i \in {1,\cdots,p}$. Now, if we apply $T$, follows that:
$$a_1v_1+\cdots+a_pv_p =0  \Rightarrow T(a_1v_1+\cdots+a_pv_p)=T(0)$$
But, $T$ is linear, and so $T(0)=0$. Thus,
$$a_1T(a_1)+\cdots+a_pT(v_p)=0$$
that is a linear combination of vectors $T(v_1), \cdots, T(v_p)$, such that at least one coefficient is non null. This means that the set $\{T(v_1),\ldots, T(v_p)\}$ is linearly dependent.
A: This is a direct consequence of linearity. I mean, it's almost by definition. Independence can be established by looking at a single equation- $c_1v_1 + \cdots + c_nv_n = 0.$ Applying any linear transformation to that equation wont affect its solutions.
