How Do Linear Transformations Affect One-to-One and Onto Properties?

[bifodus]

Let S:V --> W and T:U --> V be linear transformations. Prove that
a) if S(T) is one-to-one, then T is one-to-one
b) if S(T) is onto, then S is onto

This makes intuitive sense to me, since S(T) maps U to W, but I can't figure out how to go about proving this.

I would appreciate any help at all. Thank you.

 

[arildno]

a) Suppose that T is not one-to-one, but S(T) is one-to one.
Since there exist at least u1, u2 in U, so that T[u1]=T[u2]=v1,
then we would have S(T[u1])=S[v1]=S(T[u2])=w1, i.e, we have a contradiction, since S(T) is premised to be one-to-one.

 

[wais]

a) To prove that if S(T) is one-to-one, then T is one-to-one, we will use a proof by contradiction. Assume that T is not one-to-one. This means that there exist two distinct vectors u1 and u2 in U such that T(u1) = T(u2). Since S is a linear transformation, we have S(T(u1)) = S(T(u2)). But since S(T) is one-to-one, this implies that u1 and u2 must be equal, which contradicts our initial assumption that they are distinct. Therefore, T must be one-to-one.

b) To prove that if S(T) is onto, then S is onto, we will again use a proof by contradiction. Assume that S is not onto. This means that there exists some vector w in W that is not in the range of S. Since S(T) is onto, there must exist some vector u in U such that S(T(u)) = w. But this implies that w is in the range of S, which contradicts our initial assumption. Therefore, S must be onto.