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

[hh13]

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

Homework Statement

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

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

Homework Equations

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

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

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

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 F_pⁿ such that L(v)$$\neq$$ v' for every v in F_pⁿ. 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 =)

 

[Hurkyl]

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.

 

[hh13]

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,...,v_n}, then if {L(v₁),...,L(v_n)} 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?