1-1 and onto linear transformation question

[ironman1478]

Homework Statement

Assume that T:C^0[-1,1] ---> ℝ. assume that T is a linear transformation that maps from the set of all continuous functions to the set of real numbers.
T(f(x)) = ∫f(x)dx from -1 to 1. is T one to one, is it onto, is it both or is it neither.

Homework Equations

definition:
one to one: if v(1) != v(2) then T(v(1)) != T(v(2)). the numbers in parenthesis are subscripts.
W
onto: if the range (image) of T is the whole of W; if every w ε W is the image under T of at least one vector v ε V

The Attempt at a Solution

i was able to show that it was not one to one because there are multiple values of f(x) that map to the same real number. however i am not sure if my justification for it being onto is correct. i said that because the dimension of C^0[-1,1] is infinite and the dimension of ℝ is one, we can say that dim[V] > dim[W] which implies that T is onto.

there is a little corollary to a theorem in my book that says that if T is onto, dim[V] >= dim[W]. because i showed that dim[V] > dim[W], can i say that T is onto?

 

[Fredrik]

i am not sure if my justification for it being onto is correct. i said that because the dimension of C^0[-1,1] is infinite and the dimension of ℝ is one, we can say that dim[V] > dim[W] which implies that T is onto.

This argument is incomplete at best. Consider e.g. the map P:ℝ³→ℝ² defined by P(x,y,z)=(0,z). Clearly the dimensions of the domain and codomain are 3 and 2 respectively, but P is not surjective (onto). This P is also not injective. Your argument might work for injective functions, but I'm too tired to think about that right now.

It's very easy to prove that your T is surjective directly from the definition. Just let r be an arbitrary real number, and find a continuous function f such that Tf=r.