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## Homework Statement

Given:

T is a transformation from V -> W and the dim(V) = n and dim(W) = m (I think the dimensions were given for the purposes of another problem)

Prove:

If T is 1 to 1 and {v1, ..., vk} is a subspace of V that is L.I., then {T(v1), ..., T(vk)} is L.I in W.

## The Attempt at a Solution

First we form the set of transformations

A = {T(v1), ..., T(vk)}.

Since v1, ..., vk are L.I., none of these vectors are equal to each other.

Then since T is 1 to 1, T(vi) =/= T(vj) where vi and vj represent any vectors from {v1, ..., vk}.

Finally, this means that c1T(v1) + ... + ckT(vk) = 0 has only the trivial solution c1...ck = 0.

Therefore, {T(v1), ..., T(vk)} is L.I in W.