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Given a transformation T:V-W that is injective, if {v1,v2,...,vn} is a linearly independent subset of V, I need to prove that the set {T(v1), T(v2),...,T(vn)} will be a linearly independent set of W.

To my understanding no values in {v1,v2,...,vn} are equal since they are independent and because there is a one to one mapping from V to W the set {T(v1), T(v2),...,T(vn)} should also have no equal values. This somewhat proves that {T(v1), T(v2),...,T(vn)} is independent.

I am struggling to understand how to prove that the set {T(v1), T(v2),...,T(vn)} doesn't contain values that are linear combinations of other values in the set.

Nikki
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  • try to go with the definition of linear independence, and use the properties of a linear map – Exodd Jun 26 '20 at 11:59
  • From what I can gather from the linked question from @KaviRamaMurthy is that if the set {T(v1), T(v2),...,T(vn)} is linearly independent there exists only one solution to the equation c1T(v1)+⋯+cnT(vn)=0 and that is the zero vector. c1T(v1)+⋯+cnT(vn) = T(c1v1+...+cnvn) and since (v1,...,vn) is linearly independent c1v1+...+cnvn = 0 and T(0) = 0 (by definition of a linear map) therefore {T(v1), T(v2),...,T(vn)} is linearly independent? – Nikki Jun 26 '20 at 12:22

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