# Onto equivalent to one-to-one in linear transformations

1. Sep 23, 2015

### HomogenousCow

Can't quite see why a one-to-one linear transformation is also onto, anyone?

2. Sep 23, 2015

### WWGD

Rank Nullity Theorem: Nullity is zero....

3. Sep 23, 2015

### mathwonk

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. Sep 24, 2015

### FireGarden

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

$\dim(V)>\dim(W)$: No map is injective (one-to-one), but they can be surjective (onto). Example; $\mathbb{R}^2\rightarrow \mathbb{R},\ (a,b)\mapsto a+b$
$\dim(V)<\dim(W)$: No map is surjective, but they can be injective. Example; $\mathbb{R}\rightarrow\mathbb{R}^2,\ x \mapsto (x,0)$
$\dim(V)=\dim(W)$: 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. Sep 24, 2015

### WWGD

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.

Last edited: Sep 24, 2015