Linear applications simple doubt

[TrickyDicky]

Is an injective endomorfism automatically bijective?

 

[R136a1]

If you are talking about linear maps between finite-dimensional spaces, then yes. This is known as the Fredholm alternative: http://en.wikipedia.org/wiki/Fredholm_alternative

 

[TrickyDicky]

If you are talking about linear maps between finite-dimensional spaces, then yes. This is known as the Fredholm alternative: http://en.wikipedia.org/wiki/Fredholm_alternative

Thanks for confirming that intuition and for the pointer to Fredholm alternative, never heard of it.

 

[WWGD]

Doesn't this follow from rank-nullity theorem?

 

[mathwonk]

for finitely generated modules over other rings this can fail, e.g. if Z is the integers, the injective endomorphism Z-->Z taking n to 3n is injective but not surjective. Interestingly however, a surjective endomorphism of a finitely generated module is always injective as well.

 

[R136a1]

Doesn't this follow from rank-nullity theorem?

Yes. But the Fredholm alternative also hold in some infinite-dimensional cases. There, you can't use rank-nullity.