# Homework Help: Field of modulo p equiv classes, how injective linear map -> surjectivity

1. Nov 24, 2009

### hh13

Field of modulo p equiv classes, how injective linear map --> surjectivity

1. The problem statement, all variables and given/known data

Let Fp be the field of modulo p equivalence classes on Z. Recall that |Fp| = p.

Let L: Fpn-->Fpn be a linear map. Prove that L is injective if and only if L is surjective.

2. Relevant equations

|Fpn| = pn. We also know that for an F-linear map L: V-->V, that L is onto iff {v1,...,vm} spans V, given that DimV=m<infinity and that {v1,...,vm} are column vectors of the mxm matrix associated with L.

And that L is injective iff {v1,...,vm} is linearly independent.

And that L is bijective iff {v1,...,vm} is a basis.

3. The attempt at a solution

I just don't know what it means for a linear map to have the field of modulo p equivalence classes. on Z, let alone try and do this proof. If we suppose L is injective, then kerL=0. If L is NOT surjective, then there exists some v' in Fpn such that L(v)$$\neq$$ v' for every v in Fpn. I believe this has something to do with the field of modulo p equivalence classes, but again our professor hasn't really taught us those...

Thanks =)

2. Nov 24, 2009

### Hurkyl

Staff Emeritus
Re: Field of modulo p equiv classes, how injective linear map --> surjectivity

It's a trick question; the vector space structure is irrelevant!

Actually, IIRC, you could use the exact same proof that you would use for finite-dimensional real vector spaces. The field of scalars is irrelevant.

However, the fact you're working with a finite field makes things that much simpler.

3. Nov 24, 2009

### hh13

Re: Field of modulo p equiv classes, how injective linear map --> surjectivity

But how do we do that?

I'm having a hard time trying to go about proving linear maps are isomorphisms. I've heard that if, in a linear map L: V--W, if the basis for V is {v1,...,vn}, then if {L(v1),...,L(vn)} works as a basis for W, then it proves that V and W are isomorphic to each other and so L is an isomorphism...but I don't get why...or maybe I've heard wrong?