Proving a list of vectors is linearly independent if the list of its images is linearly independent This is from Axler's Linear Algebra Done Right 3rd edition, exercise #4 from 3A. I want to make sure my proof is logical.

Suppose $T\in\mathcal{L}(V,W)$ and $v_1,\dots,v_m$ is a list of
  vectors in $V$ such that $Tv_1\dots,Tv_m$ is linearly independent in $W$.
  Prove that $v_1,\dots,v_m$ is linearly independent.
Proof. Since $Tv_1\dots,Tv_m$ is linearly independent, then $c_1Tv_1+\dots+c_mTv_m=0\implies c_j=0$ for $j=1,\dots,m$. Since $T$
  is linear, we have $T(c_1v_1+\dots+c_mv_m)=T(0)$, so
  $c_1v_1+\dots+c_mv_m=0$ where each $c_j=0$ from above.

I'm curious because I'm not sure if this guarantees that since each $c_j=0$ in this particular case (the transformations being linearly independent) means that $\sum^m_{i=1}c_iv_i$ must be $0$.
 A: If your goal is to show (directly) that the $v_i$ are linearly independent, the proof should really be structured as 

  
*
  
*Assume that the linear combination $c_1v_1 + \ldots c_mv_m = 0$. Then...
  
*???
  
*Profit! Therefore $c_1 = \ldots = c_m = 0$,
  

where step 2. somehow uses the linear independence of the $Tv_i$.
Unfortunately, yours seems to have shown that whenever each of the $c_i$ are zero, the linear combination $c_1v_1 + \ldots c_mv_m$ is the zero vector.
Note: There may be other issues. It's hard to say, but you also seem to assume that if $T(c_1v_1 + \ldots c_mv_m) = T(0)$, then $c_1v_1 + \ldots c_mv_m = 0$, which really only works if $T$ is injective. However, $T$ doesn't need to be; the proof is true for arbitrary $T$. This amounts, in a certain sense, to trying to bring information from $W$ back to $V$, which isn't possible given a generic transformation $T\colon V \to W$ (but would be possible for injective $T$!). 
Essentially, you'll want to use $T\colon V \to W$ to push information about the coefficients $c_i$ forward to $W$, and then draw conclusions from the $Tv_i$.
So, you really just need to rearrange your proof and be a bit more careful (about wording, if nothing else). 


*

*Assume $\sum_i c_iv_i = 0$.

*Now apply $T$, so that $T\left(\sum_i c_i v_i\right) = \sum_i c_iT(v_i) = 0$.

*Profit! Therefore $c_i = 0$ for all $i$.

