Question 12549
  Linear operators are not merely functions on sets,they are linear transfomrations(preserving addition and scalar multiplication of vectors on vector space. 

 I feel very strange that you haven't mentioned vector space, or matrices.
 Without them, how can you start solving the questions ?

 1) I'm looking for a linear operator from V->V such that T^2=0 but T does not =0. 
  Sol: Define {{{ T:R^2}}} --> {{{R^2 }}} generated by
   T(1,0) = (0,1) and T(0,1) = (0,0)
 [Note : (1,0) & (0,1) are standard basis of unitcolumn vectors]
  Let B = {(1,0),(0,1)} 
 The matrix A = [T}B of T associated with the basis B as
 [0 0] 
 [1 0] 
 or equivalently T(X) = AX for all column vector X in {{{R^2}}}
 Clearly,you can see that {{{A^2 = 0 }}} but A is not 0.
  More precisely, T(a, b) = aT(1,0) + bT(0,1) = a(0,1) = (0,a)
 for all (a,b) in {{{ R^2}}}
 Check: Clearly,T is not the zero operator (why?) and
 we see that {{{ T^2}}}(a,b) = T(0,a) = 0*T(1,0) = (0,0) for all real a,b
 2) Given two linear operators (say, T, U) from V-> V, I'm looking for TU=0 but UT does not =0
 Sol: Similarly to the example in 1)
      Let T ,U be two linear operators in {{{R^2}}} defined by
     T = (as matrix A) 
    [0 1]
    [0 0] and
    U = (as matrix B)
    [1 0]
    [0 0] then we have
    TU =
    [0 0]
    [0 0] but UT =
    [0 1]
    [0 0] (not zero operator)
  More precisely, define {{{ T:R^2 }}} --> {{{R^2 }}} by T(X) = AX 
  and {{{ T:R^2}}} --> {{{R^2 }}} by T(X) = BX where X is any column vector of {{{R^2}}}.
  AX & BX are products of matrices.

 Make sure you do understand the examples above and work hard. 

 Kenny