Linear Algebra: basic proof about a linear transformation

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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.
 

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  • #2
vela
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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.
{v1, …, vk} isn't a subspace, and it doesn't make sense to say a subspace is linearly independent. {v1, …, vk} is simply a set of k linearly independent vectors in V.

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.
Just because two vectors aren't equal to each other doesn't mean they're linearly independent. The vectors (1,0,0) and (2,0,0) for example aren't equal to each other, but they are dependent.
Therefore, {T(v1), ..., T(vk)} is L.I in W.
You want to use the fact that T is linear. (You didn't say it was, but I assume that it's a given in the problem.) What can you say about the null space of T since it's one-to-one?
 
  • #3
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Thanks vela. Sorry about mis-wording the question, and yes, T is linear. Here is the revised version of my proof:


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

For {T(v1), ..., T(vk)} to be L.I., we must have the trivial solution for c1T(v1) + ... + ckT(vk) = 0 . Or, by linearity, T(c1v1 + ... + ckvk) = 0 .

Since T is 1 to 1, the null space of T must be 0. That is, T(0) = 0. This means that c1v1 + ... + ckvk = 0.

Because v1, ..., vk are L.I., all the c's must equal 0 (aka we have only the trivial solution).

Therefore we can conclude that {T(v1), ..., T(vk)} is L.I in W.
 
  • #4
vela
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Perfect!
 

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