Linear Algebra Composition Isomorphism Question

[Nexttime35]

Homework Statement

Let S : U →V and T : V →W be linear maps.

Given that dim(U) = 2, dim(V ) = 1, and dim(W) = 2, could T composed of S be an isomorphism?

Homework Equations

If Dim(v) > dim(W), then T is 1-1
If Dimv < dim(w), then T is not onto.

The Attempt at a Solution

So this seems like a tricky question, but I am having trouble proving whether or not this is an isomorphism. While I know that T composed of S (u) = T(S(u)), S: U→V and T:V→W, this would be an isomorphism if dim(U) = Dim(W). But does the fact that S maps to V, which is one dimension less than U and W, affect that this is an isomorphism?

I think an example could be helpful in helping me understand this problem.
Thank you.

 

[kduna]

If Dim(v) > dim(W), then T is 1-1

That should read "If Dim(V) > Dim(W), then T is not 1-1". I'm sure that's what you meant, but just making sure.

Anyway, let's suppose that TS is an isomorphism. Which would mean that given w \in W, there is a u \in U. Such that TS(u) = w. Is this a problem?

 

[Nexttime35]

That makes sense, kduna, thank you. So Basically this composition bypasses the use of V, and there exists an isomorphism from U to W. Thanks.

 

[kduna]

That makes sense, kduna, thank you. So Basically this composition bypasses the use of V, and there exists an isomorphism from U to W. Thanks.

Actually no. w = TS(u) = T(S(u)). So w \in Range(T). But this can be done for any w \in W. This would mean T is surjective, a contradiction since dim(V) < dim(W).

Therefore TS cannot be an isomorphism.

Note that since dim(U) = 2 = dim(W), an isomorphism does exist between U and W. But TS is not an isomorphism.

 

[Nexttime35]

Ahh, I see. That makes much more sense. Thank you for your help, kduna.