Question 1031968
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 {{{ c[1]X[1] + c[2]X[2]}}}+...+{{{c[n]X[n] = theta[V]}}}, where {{{theta[V]}}} is the zero vector in V.
Then linear independence of the set implies that only {{{c[1] = c[2] = c[3]}}} = ...= {{{c[n-1] = c[n] = 0}}} will satisfy the previous equation.


Now let  
{{{d[1]L(X[1]) + d[2]L(X[2])}}}+...+{{{d[n]L(X[n])}}} ={{{ theta[W]}}} <-----Equation A.               
({{{theta[W]}}} 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 

{{{L(d[1]X[1]) + L(d[2]X[2])}}}+...+{{{L(d[n]X[n])}}}) ={{{ theta[W]}}}, or 

L({{{d[1]X[1]+ d[2]X[2]}}}+...+{{{d[n]X[n]}}}) ={{{ theta[W]}}}.

==> {{{d[1]X[1]+ d[2]X[2]}}}+...+{{{d[n]X[n]}}} = {{{theta[V]}}},

or the left-hand side linear combination would be an element of the kernel of L.

==>  {{{d[1]= d[2]}}}=...={{{d[n]=0}}}, 

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.