Originally Posted by

**jsndacruz** The problem assumes V and W are vector spaces, and that T: V --> W is linear. We must prove that T is one to one if and only if T carries linearly independent subsets of V onto linearly independent subsets of W;

My attempt:

I first assume T one to one. Then N(T) = {0}. Assume V is linearly independent, but its image in W was not. Then:

$\displaystyle \sum_{i=1}^{n} a_i w_i = 0$. But $\displaystyle \sum_{i=1}^{n} a_i w_i = \sum_{i=1}^{n} a_i T(v_i) = T \sum_{i=1}^{n} a_i v_i$. Since T is one to one, an inverse exists, so we get :

$\displaystyle T^{-1} T \sum_{i=1}^{n} a_i w_i = \sum_{i=1}^{n} a_i v_i = T^{-1} (0) = 0.$ But then we have a linear combination of vectors in v that sum to the zero vector, contradicting our assumption that V was linearly independent.

I'm not sure if this part is right, and I'm also not sure how to prove going the other way (that if it carries linear independence, it's one to one).