- #1
brushman
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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.