Solved] Proving Linear Transformation Properties with Linear Independent Sets

[amanda_ou812]

Homework Statement

Let V and W be vector spaces and T: V-> W be linear.
a) Prove that T is one to one if and only if T carries linearly independent subsets of V onto linearly independent subsets of W.
b) Suppose that T is one to one and that S is a subset of V. Prove that S is linearly independent if and only if T(S) is linearly independent.
c) Suppose B = {v1, v2, ..., vn} is a basis for V and T is one to one and onto. Prove that T(B) = {T(v1), ..., T(vn)} is a basis for W.

Homework Equations

1. nullity + rank = dim V
2. T is one to one if and only if N(T) = {0}
3. T is onto implies R(T) = W
4. R(T) = span (T(B))
5. there may be others

The Attempt at a Solution

a) If T is one to one then N(T)) = {0} which implies that nullity = 0. Then we know that rank = dim V. This implies that T is onto (by a theorem in my text). This means that R(T) = W = span (T(B)) where B is a basis for V. So W = a T(v1) + ...+ a (sub n) T(vn) where B = {v1, ..., vn}. HERE is where I think I NEED another step before I can do the conclusion... So this shows that T carries linearly independent subsets of V onto linearly independent subsets of W

If T carries linearly independent subsets of V onto linearly independent subsets of W, then T is onto, which implies that R(T) = W. this implies that dim R(T) = rank = dim W which implies that rank = dim V and therefore T is one to one.

b) I do not have anything formal...just my thought process
So, I know T is one to one, S is a subset of V and S is linearly independent. So T is one to one implies that N(T) = {0} which implies that rank = dim V which implies that T is onto which implies that R(T) = W = span (T(B)) where B is a basis for V. So if S = B then we know that R(T) = W = span (T(S)). I know that S is linearly independent ... but how do i get to T(S) being linearly independent.

For the other way, I know T is one to one, S is a subset of V and T(S) is linearly independent. I know the same stuff as above but I do not know how T(S) linearly independent implies S is linearly independent.

c)
So, same thoughts here. T is one to one implies that N(T) = {0} which implies that rank = dim V. Also, T is onto which implies that R(T) = W = span (T(B)) where B is a basis for V. ok, so how do I show that it is a basis for W?
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[Ryker]

For c), suppose not. Suppose there is a vector T(v_x), which can be expressed in terms of other T(v_i). Knowing T is linear, what does that imply?

 

[amanda_ou812]

Then T is linearly dependent. So there exists lambda's, not all 0, such that the linear combination is 0. Also, if T is linearly dependent than v1, ..., vn are linearly dependent so then B is not a basis for V.