# Thread: Linear Transformations and Linear Independence

1. ## Linear Transformations and Linear Independence

Let L: V -> W be a linear transformation and let S = {v1, ... vn} be a set of vectors in V. Prove that if T = {L(v1), ... L(vn)} is linearly independent, then so is S. What can we say about the converse?

So I gave the problem a try. Not sure if I got it right or if I did it backwards! Having trouble with the converse though, so feedback is greatly appreciated!

Let's say L is one to one.

Let a1L(v1) + ... + anL(vn) = 0w where a1, ... an are real numbers

L(a1v1 + ... + anvn) = 0w = L(0v)

Since L is one to one we conclude that a1v1 + ... + anvn = 0v

This would imply that S is linearly independent because a1, ... an = 0.

I'm not sure how I'd start with the converse. That would be that S is linearly independent so then T would be linearly independent. I got stuck there. Thanks guys!

2. Originally Posted by Brokescholar
Let L: V -> W be a linear transformation and let S = {v1, ... vn} be a set of vectors in V. Prove that if T = {L(v1), ... L(vn)} is linearly independent, then so is S. What can we say about the converse?

Let's say L is one to one.
No need to make this assumption. Say that $\{\bold{v}_1,...,\bold{v}_n\}$ is not linear independent then there are $a_i$ not all zero so that $\sum_i a_i \bold{v}_i = \bold{0}$ but then $L\left( \sum_i a_i \bold{v}_i \right) = L(\bold{0}) = \bold{0}$. But $L$ is linear and so $\sum_i a_i L(\bold{v}_i) = \bold{0}$. Therefore, $\{ L(\bold{v}_1),...,L(\bold{v}_n)\}$ is not linearly independet. Now take the contrapositive of this statement to complete the proof.

The converse is not true unless we know that $L$ is a one-to-one linear transformation.
Can you come up with a conterexample?