Question 18994
 a linear transformation from {{{R^3}}} to {{{R^2 }}}?
 L(x1,x2,x3) = (x2-x3,x1+2) ? 
 (i) It is still closed under scalar multiplication and addition  but
 not linear. Simly, because L(0,0,0) != (0,0).

 Since, for any   a linear transformation L, L(0) = L(0+0)= L(0) + L(0)
 so L(0) = 0 for zero vector. 

 In general, such L {{{R^m}}} to {{{R^n }}} if and only if each n component
 of L is a linear function. Now the 2nd component x1+2 is not linear.
 [You idea was Ok with wrong reason.]


so i know how to prove that this is not closed under scalar multiplication and addition (due to the "2" in the second entry). so then is my answer just "no, it is not a linear transformation from R^3 to R^2 because it is not closed   
under scalar multiplication." ? 

 NO !!!!

what about linear transformation from R^3 to R^4: L(x1,x2,x3) = (x2-x3,x1*x2,2x2-x1,x3+x2) ? 

 No, since the 2nd component x1*x2 is not linear.
 More precisely,
 L(x1,x2,x3) = (x2-x3,x1*x2,2x2-x1,x3+x2) 
 
 Since (1,1,0) = (1,0,0) +(0,1,0).
 L(1,1,0) = (1, 1,1, 1)
 L(1,0,0) = (0, 0, -1, 0) and L(0,1,0) = (1, 0, 2, 1) 

 but L(1,0,0) + L(0,1,0) = (1,0,1,1) != L(1,1,0)

 Hence,L is not a linear transformation.

 Another wrong idea:
" if{{{ R^m}}} ---> {{{R^n}}}, then m must be greater than or equal to n. "

 No, m ,n can be any two non-negative integers, not restricting to m<=n.

 Good luck !

 Kenny