If this problem talked about "[itex]R^3\to R^3[/itex]" (or, more generally, [itex]R^n\to R^n[/itex] with n odd) the answer would be, very quickly "no" because of the "rank-nullity" theorem- the rank and nullity of a linear operator must add up to the dimension of the range space. If Ker(L)= Range(L), they would have to have the same dimension and so the sum of dimensions would be an even number, not odd.

That also tells us that, in this case, with dimension 4, we must look for a linear transformation whose kernel has dimension 2, and whose range has dimension 2- so that 2+ 2= 4. What would be the simplest mapping of <a, b, c, d> into a four-vector with kernel of dimension 2?