Linear Algebra: Linear Transformation proof Let T: V->W be a linear transformation between vector spaces over F and let $v_1,v_2...,v_n$ elements of V
if ${Tv_1, Tv_2, ..., Tv_n}$ is linearly independent, prove that ${v_1,v_2,...,v_n}$ linearly independent as well 
So far I have, 
$a_1Tv_1 + ... + a_nTv_n = 0$ where $a_1,...,a_n = 0$ 
then, $T(a_1v_1 + ...+ a_nv_n) = 0$
is it enough now to say that 
$a_1v_1 + ... + a_nv_n = 0$ and since ${a_1,...,a_n}=0$ 
${v_1,v_2,...,v_n}$ linearly independent
 A: This is wrong. You can't assume that the $a_n$'s are $0$. You are supposed to prove that!
Take scalars $a_1,\ldots,a_n$ such that $a_1v_1+\cdots+a_nv_n=0$; you want to prove that each $a_i$ is $0$. But\begin{align}a_1v_1+\cdots+a_nv_n=0\implies&T(a_1v_1+\cdots+a_nv_n)=0\\\iff&a_1T(v_1)+\cdots+a_nT(v_n)=0\\\implies&a_1=\cdots=a_n=0,\end{align}since $T(v_1),\ldots,T(v_n)$ are linearly independent.
A: We want to prove that $v_1,\dots,v_n$ are independent, so start out from their combination instead: Suppose
$$a_1v_1+\dots+a_nv_n=0$$

 Now apply $T$ on both sides, use its linearity, and then use the condition, in order to conclude that all $a_i=0$. 

A: Actually, you wrote things backward. Why did you take $a_1Tv_1+\dots+a_nTv_n=0$ with $a_1,\dots,a_n=0$? Actually, I can really take a linearly dependent family of vectors $v_1=v_2=\dots=v_n=v$ and say "take $a_1v_1+\dots+a_nv_n=0$ with $a_1=\dots=a_n=0$. Then the family is independent because $a_i=0$".
The casual way you prove $P\implies Q$ is that you assume $P$ is true and prove $Q$.
You want to prove that

$Tv_1,\dots,Tv_n$ are linearly independent $\implies$ $v_1,\dots,v_n$ are linearly independent.

So, let's go. Assume that $Tv_1,\dots,Tv_n$ are linearly independent. You tried to translate what it means, which is a good thing, but used it in a quite wrong way. Your assumption means that:

If $a_1,\dots,a_n$ are real numbers such that $a_1Tv_1+\dots+a_nTv_n=0$, then $a_1=\dots=a_n=0$.

This is your assumption. This is a hypothesis. You have nothing to use it at for the moment.
Now what's your goal? You want to show that $v_1,\dots,v_n$ are linearly independent. Translate this:

If $a_1,\dots,a_n$ are real numbers such that $a_1v_1+\dots+a_nv_n=0$, then $a_1=\dots=a_n=0$.

So let's prove this. Let $a_1,\dots,a_n$ be real numbers such that $a_1v_1+\dots+a_nv_n=0$. Why is $a_1=\dots=a_n=0$? I'll let you find out.
A: No. You want $a_1\cdot v_1+\dots+a_n\cdot v_n=0 \implies a_1=a_2=\dots=a_n=0$.
Now, $T(0)=0$, since $T$ is linear.  Also by linearity,  $a_1T(v_1)+\dots+a_nT(v_n)=0$. 
Now conclude $a_1=a_2=\dots =a_n=0$ (by linear independence of $T(v_1),\dots ,T(v_n)$...)
