1. ## Transform one-to-one

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

2. ## Re: Transform one-to-one

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:
$\sum_{i=1}^{n} a_i w_i = 0$. But $\sum_{i=1}^{n} a_i w_i = \sum_{i=1}^{n} a_i T(v_i) = T \sum_{i=1}^{n} a_i v_i$. Since T is one to one, an inverse exists, so we get :
$T^{-1} T \sum_{i=1}^{n} a_i w_i = \sum_{i=1}^{n} a_i v_i = T^{-1} (0) = 0.$ 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).
You omitted a lot of steps...and words...but yeah, that's the right idea. To do the converse, why don't you try thinking about what would happen if $\displaystyle \sum_i a_i v_i$ (where, presumably as you mean, $\{v_1,\cdots,v_n\}$ is a basis for [tex]V[tex]) mapped to zero? This would tell us that $\displaystyle \sum_i a_i T(v_i)=0$, but what did we just assume about $T$, that tells us about $\{T(v_1),\cdots,T(v_n)\}=T(\{v_1,\cdots,v_n\})$. In turn, what does this tell us about $\displaystyle \sum_i a_i v_i$, and moreover what does it tells us about $\ker T$.

3. ## Re: Transform one-to-one

Sorry for the sloppy derivation - was just trying to sketch out the proof first. And you were right in thinking I meant {v_1..v_n} was a basis. Thanks a lot!