First, it follows from the rank-nullity theorem that any complementary subspace for R must have the same dimension as N.
So if then it's obvious that N is a T-invariant complementary subspace for R.
Conversely, suppose that S is a T-invariant complementary subspace for R. If then (obviously, since Tx is in the range of T) and (because S is T-invariant). Therefore . In other words, . Hence . But these subspaces have the same dimension, and so S = N.