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## Main Question or Discussion Point

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

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

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WWGD

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Rank Nullity Theorem: Nullity is zero....Can't quite see why a one-to-one linear transformation is also onto, anyone?

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mathwonk

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In general they aren't. If the transformation is [itex]V\rightarrow W[/itex] with V,W finite dimensional, then there are three cases:Can't quite see why a one-to-one linear transformation is also onto, anyone?

[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.

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WWGD

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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.Rank Nullity Theorem: Nullity is zero....

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