1. ## linear independence

how do you prove that if {f(v1), f(v2),...f(vn) } is linearly independent where f:V to W then f is injective?

by def of LI, a linear combination of f(v1), f(v2),...f(vn) = 0 iff the coefficients are 0. so can i say that for all w in W, there is a unique linear combi of {f(v1), f(v2),...f(vn) } which will give w.
this means that w in W is unique...

and then how do i continue?

2. Prove that {v_1,...,v_n} is linearly independent also and use the following theorem:

Let V and U be vector space over field K. Let {v_1,...,v_n} a basis of V and {u_1,...,u_n} a basis of U. Then exist a unique linear transformation F:V-->U which:

F(v_1)=u_1, ... , F(v_n)=u_n.

3. Why are we able to use the second theorem? I thought by saying that there is a unique linear transformation, we are assuming that it is injective.. Which is something we need to prove.

Thanks!

4. Originally Posted by alexandrabel90
how do you prove that if {f(v1), f(v2),...f(vn) } is linearly independent where f:V to W then f is injective?
You are leaving out a lot here. In order for this to be true, {v1, v2, ..., vn} has to be a basis for V.

by def of LI, a linear combination of f(v1), f(v2),...f(vn) = 0 iff the coefficients are 0. so can i say that for all w in W, there is a unique linear combi of {f(v1), f(v2),...f(vn) } which will give w.
this means that w in W is unique...

and then how do i continue?
Let u and w be such that f(u)= f(v). Then, since {v1, v2,..., vn} is a basis for V, there exist a1, a2, ..., an such that a1v1+ a2v2+ ...+ anvn= u and b1, b2, ..., bn such that b1v1+ b2v2+ ...+ bnvn= v. Then, since f(u)+ f(v), a1f(v1)+ a2f(v2)kl+ ...+ anf(vn)= b1f(v1)+ a2f(v2)+ ...+ bnf(vn). Rewrite that so that you have a linear combination of f(v1), f(v2), ..., f(vn) equal to 0 and use the fact that they are independent to show that a1= b1, a2= b2, ..., an= bn and so u= v.

5. Opps ya. v1 to vn is a basis for V. Sorry!