Question 8159
 Yes, since (1,-1,1) and (1,1,1) are linearly independent  in R^3
 Also, (1,0) and (0,1) are linearly independent in R^2 can form a basis.

 We know that any linear transformation is uniquely determined by
 the the values on the basis.
 Set another vector v in R^3, which is independent of (1,-1,1) and (1,1,1),
 say (0, 0,1) , then define T(0,0,1)= (0,0) [or choose v = (1,-1,1)x(1,1,1)

 We obtain a linear transformation from R^3 to R^2 generated by 
 T(1,-1,1)=(1,0)
 T(1,1,1)=(0,1) with Kernel(T) = <(0,0,1)>
 
 More precisely, T(a(1,-1,1)+b(1,1,1)+c(0,0,1))
 = a(1,0)+ b(0,1)= (a,b) for all real a,b,c

 Kenny
 PS: Important notice:
 I will not solve your or other student's questions with repeated posting.