Linear Algebra Linear maps dealing with linear independence.

Hi, I have this proof but I just want to know if I'm doing this right, or if there's a better way to write my proof (which is incomplete).

5. Let V and W be vector spaces over F, and suppose that T in L(V,W) is injective. Given a linearly independent list (v1, ..., vn) of vectors in V, prove that the list (T(v1), ..., T(vn)) is linearly independent in W.

So, I think the idea behind proving this is that, injective means that every element of dom(T) has a unique mapping to range(T), thus when you have a list of vectors of the mapping of every vector in the (v1, ... vn), each mapping will be unique, and thus that list in linearly independent?

I'm a bit all over the place, and I guess i'm just having trouble finding a way to put this, and also wondering if I'm missing anything important.

Re: Linear Algebra Linear maps dealing with linear independence.

Adding more thought, what exactly would imply that (w1, ..., wn), from T(v1, ..., vn)=(w1, ..., wn) for a vector in W, is linearly independent too from the fact that (v1, ..., vn) is linear independent and T is injective?

Re: Linear Algebra Linear maps dealing with linear independence.

I would prove it by contraposition.

Suppose is linearly dependent then we could for example write as a linear combination of the other vectors, say

with . Since is a linear map we can write , as is injective we have but that means is linearly dependent.

Re: Linear Algebra Linear maps dealing with linear independence.

Thank you but the proof was to prove that {T(v1), ..., T(vn)} was linearly Independent.

Re: Linear Algebra Linear maps dealing with linear independence.

Quote:

Originally Posted by

**zachoon** Thank you but the proof was to prove that {T(v1), ..., T(vn)} was linearly Independent.

Yes, that's the reason I said I would give a proof by contraposition.