Proving One-to-One Property of Linear Transformations with Dimension Equality

[baileyyc]

I am having trouble with this problem:
Let T:V->W be a linear transformation. Prove that T is one-to-one if and only if dimension of V = dim(RangeT).

I know that in order to be a linear transformation:
1) T(vector u + vector v) = T(vector u) + T(vector v) and
2) T(c*vector u) = cT(vector u), where c is a scalar

and I know that one-to-one means that there is exactly one x for every y, and that there are no unmapped elements.

and that the dimension of a vector space is the number of vectors in a basis.

But I'm not sure what I'm actually proving..

 

[jbunniii]

Do you know a theorem that relates the dimensions of V, range(T), and kernel(T)?

Hint: T is one-to-one if and only if kernel(T) = {0}.

(kernel means the same thing as null space, in case you haven't seen the term)

 

[fzero]

What if there is are non-zero vectors k_i such that T(k_i)=0?. What can you say about T(c k_i) and dim(range(T))? Can you relate dim V to dim(range(T)) and the number of linearly independent such vectors k_i?