Onto equivalent to one-to-one in linear transformations

[HomogenousCow]

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

 

[WWGD]

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

Rank Nullity Theorem: Nullity is zero...

 

[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.

 

[FireGarden]

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

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.

 

[WWGD]

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.