1-1 and onto linear transformation question

In summary: By the fundamental theorem of calculus, this is not difficult at all.In summary, T is a linear transformation from the set of continuous functions to the set of real numbers, defined as the integral from -1 to 1 of a continuous function. It is not one to one as there are multiple values of f(x) that map to the same real number. However, it is onto as for any real number r, there exists a continuous function f such that Tf=r, proven by the fundamental theorem of calculus.
  • #1
ironman1478
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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?
 
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  • #2
ironman1478 said:
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.
 

Related to 1-1 and onto linear transformation question

1. What is a 1-1 linear transformation?

A 1-1 linear transformation, also known as an injective linear transformation, is a mathematical function that maps each input to a unique output. This means that no two different inputs will result in the same output. In other words, the function preserves distinctness.

2. What does it mean for a linear transformation to be onto?

A linear transformation is onto, or surjective, if every element in the output space is mapped to by at least one element in the input space. In other words, the function covers or reaches all elements in the output space.

3. How can I determine if a linear transformation is 1-1 and onto?

In order for a linear transformation to be both 1-1 and onto, it must satisfy two conditions: 1) for every distinct pair of elements in the input space, the resulting outputs must also be distinct, and 2) every element in the output space must be mapped to by at least one element in the input space.

4. What is the importance of 1-1 and onto linear transformations?

1-1 and onto linear transformations are important in mathematics and science because they preserve the structure and properties of the original space. They are also useful in solving equations and understanding the behavior of systems.

5. How can I use 1-1 and onto linear transformations in real life?

1-1 and onto linear transformations have many applications in real life, such as in computer graphics, data compression, and signal processing. They are also used in cryptography, where 1-1 and onto functions are used to encrypt and decrypt data.

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