SOLUTION: Let L:V→W be a linear transformation. Let {(X1),(X2),, (Xn)} ϵ V. IF {L(X1),L(X2),, L(Xn)} is linearly dependent, then {X1,X2,...,Xn} is linearly dependent.

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Question 1031968: Let L:V→W be a linear transformation. Let {(X1),(X2),, (Xn)} ϵ V. IF {L(X1),L(X2),, L(Xn)} is linearly dependent, then {X1,X2,...,Xn} is linearly dependent.
Found 2 solutions by ikleyn, robertb:
Answer by ikleyn(52800)   (Show Source): You can put this solution on YOUR website!
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It is not necessarily true.
-----------------------------------

I just said it, and I am repeating it one more time and again.

>>> IT IS NOT NECESSARILY TRUE. <<<

It is true only in the case when the operator L is non-degenerated (has the zero kernel).
Which is not always the case for linear transformations.

The proof of the other tutor is wrong, unfortunately. 

A contr-example is:


Take 3 linearly independent vectors in .

Let the operator L be the projection  on .

Every three vectors in  are dependent.

So are dependent in  the projections of the original vectors from {{R^3}}}.

But the original vectors were chosen as linearly independent. 

It is on the level of elementary knowledge of linear algebra.

Again: the fact that the images are linearly dependent DOES NOT IMPLY that the pre-images are necessarily linearly dependent.


Answer by robertb(5830)   (Show Source): You can put this solution on YOUR website!
The problem asks to show that: If {L(X1),L(X2),, L(Xn)} is linearly dependent, then {X1,X2,...,Xn} is linearly dependent.
It would be easier to prove the contrapositive of this statement:
If {X1,X2,...,Xn} is linearly independent, then {L(X1),L(X2),, L(Xn)} is linearly independent as well. This is quite easy to prove.
DEFINITION: Let +...+, where is the zero vector in V.
Then linear independence of the set implies that only = ...= will satisfy the previous equation.

Now let
+...+ = <-----Equation A.
( is the zero vector in W and the d constants are arbitrary.)
By the property of the linear transformation L,
Equation A is equivalent to
+...+) =, or
L(+...+) =.
==> +...+ = ,
or the left-hand side linear combination would be an element of the kernel of L.
==> =...=,
by virtue of the linear independence of the X vectors.
Therefore, {L(X1),L(X2),, L(Xn)} is linearly independent as well, and the theorem is proved.


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