Function that is an isomorphism

[NATURE.M]

Homework Statement

So my text states the proposition:
If V and W are finite dimensional vector spaces, then there is an isomorphism T:V→W ⇔ dim(V)=dim(W).

So, in an example the text give the transformation T:P$_{3}$(R)→P$_{3}$(R)
defined by T(p(x)) = x dp(x)/dx.

Now I understand T is not an isomorphism since ker(T) = span(1) , the set of all constant polynomial functions. But by the above theorem since the dim(V) = dim(W), T would an isomorphism.
So I'm a bit confused.

 

[R136a1]

The theorem says that there is some function that is an isomorphism. It does not say that any function $T$ is an isomorphism.

In your example, the theorem is satisfied since $S(x) = x$ is an isomorphism. But the theorem does not say that any arbitrary $T$ is an isomorphism. So you can not deduce that your $T$ is an isomorphism from the theorem.

 

[LCKurtz]

Homework Statement

So my text states the proposition:
If V and W are finite dimensional vector spaces, then there is an isomorphism T:V→W ⇔ dim(V)=dim(W).

So, in an example the text give the transformation T:P$_{3}$(R)→P$_{3}$(R)
defined by T(p(x)) = x dp(x)/dx.

Now I understand T is not an isomorphism since ker(T) = span(1) , the set of all constant polynomial functions. But by the above theorem since the dim(V) = dim(W), T would an isomorphism.
So I'm a bit confused.

No. The theorem says that if dim(V) = dim(W) then there is an isomorphism. It doesn't say any old map you choose is an isomorphism. Your map isn't onto.

 

[NATURE.M]

Ok thanks a lot. Its makes sense now.