Onto equivalent to one-to-one in linear transformations

In summary, the Rank-Nullity Theorem states that for a linear map from a finite dimensional space to itself, if the nullity (dimension of the kernel) is zero, then the map must also be onto. This is a fundamental fact about maps of finite dimensional spaces and can be extended to linear maps between different finite dimensional spaces. However, this is not always the case, as there are instances where the map can be injective or surjective, but not both.
  • #1
HomogenousCow
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Can't quite see why a one-to-one linear transformation is also onto, anyone?
 
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  • #2
HomogenousCow said:
Can't quite see why a one-to-one linear transformation is also onto, anyone?

Rank Nullity Theorem: Nullity is zero...
 
  • #3
this is an amazing and useful fact about maps of finite dimensional spaces. it is true for maps of a finite set to itself, and the theory of dinsion of linear spaces allows us to extend it to linear maps of finiute dimensional spaces. i.e. linear injections take an n dimensional space to an n dimensional image. but if the target space is also n dimensional, then an n dimensional image subspace must fill it up, i.e. the map must be onto as well. this is a highly non trivial fact, but very fundamental.
 
  • #4
HomogenousCow said:
Can't quite see why a one-to-one linear transformation is also onto, anyone?

In general they aren't. If the transformation is [itex]V\rightarrow W[/itex] with V,W finite dimensional, then there are three cases:

[itex]\dim(V)>\dim(W)[/itex]: No map is injective (one-to-one), but they can be surjective (onto). Example; [itex] \mathbb{R}^2\rightarrow \mathbb{R},\ (a,b)\mapsto a+b [/itex]
[itex]\dim(V)<\dim(W)[/itex]: No map is surjective, but they can be injective. Example; [itex]\mathbb{R}\rightarrow\mathbb{R}^2,\ x \mapsto (x,0)[/itex]
[itex]\dim(V)=\dim(W)[/itex]: We have surjective if and only if injective.

Taking for granted every vector space has a basis, and that linear maps need only be defined on basis elements to be defined on the whole space, you can prove these by just considering maps between finite sets. This is a good exercise. You can then get more precise about this with the Rank-Nullity theorem.
 
  • #5
WWGD said:
Rank Nullity Theorem: Nullity is zero...
I guess I was assumming the same dimension for map, i.e., map from ##\mathbb R^n \rightarrow \mathbb R^n ## or any two vector spaces of the same dimension. There are other ways of seeing this. EDIT: Mayb be more accurate to say that map T is of full rank than saying it is onto.
 
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Related to Onto equivalent to one-to-one in linear transformations

1. What does it mean for a linear transformation to be onto equivalent to one-to-one?

Onto equivalence and one-to-one equivalence are two different ways to describe the same property of a linear transformation. Onto equivalence means that the transformation maps every element in the range to a unique element in the domain. One-to-one equivalence means that the transformation maps each element in the domain to a unique element in the range. Essentially, this means that the transformation is both injective and surjective.

2. What is the difference between onto equivalence and one-to-one equivalence?

The difference between onto equivalence and one-to-one equivalence lies in the perspective from which the transformation is viewed. Onto equivalence focuses on the range of the transformation, ensuring that every element in the range is mapped to a unique element in the domain. One-to-one equivalence, on the other hand, focuses on the domain of the transformation, ensuring that every element in the domain is mapped to a unique element in the range.

3. How is onto equivalence related to invertibility?

A linear transformation is invertible if and only if it is both onto and one-to-one. This means that the transformation has a unique inverse that maps the range back to the domain, and vice versa. This inverse can only exist if the transformation is onto equivalent to one-to-one, as it ensures that there are no elements in the range that are mapped to multiple elements in the domain.

4. Can a linear transformation be onto equivalent to one-to-one without being invertible?

No, a linear transformation cannot be onto equivalent to one-to-one without being invertible. As mentioned earlier, invertibility is a necessary condition for onto equivalence and one-to-one equivalence. If the transformation is not invertible, there must be either elements in the range that are not mapped to in the domain, or multiple elements in the range that are mapped to the same element in the domain, violating the requirements for onto equivalence and one-to-one equivalence.

5. How can I determine if a linear transformation is onto equivalent to one-to-one?

To determine if a linear transformation is onto equivalent to one-to-one, you can use the rank-nullity theorem. This theorem states that for a linear transformation from a vector space of dimension n to a vector space of dimension m, if the rank of the transformation is equal to n (i.e. the transformation is onto) and the nullity of the transformation is equal to 0 (i.e. the transformation is one-to-one), then the transformation is onto equivalent to one-to-one. So, you can calculate the rank and nullity of the transformation and check if they satisfy these conditions.

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