sorry wrong forum
clearly the subspace generated by the basis {(1,0,0)} is T-invariant, since that is E2.
also the subspace generated by {(0,0,1)} is T-invariant, since that is E3.
therefore the subspace generated by the basis {(1,0,0),(0,0,1)} is also T-invariant. a vector in this subspace is of the form (a,0,b),
and T(a,0,b) = (2a,0,3b) = 2a(1,0,0) + 3b(0,0,1).
two more obvious T-invariant subspaces are R^3 itself, and the 0-subspace.
a less obvious T-invariant subspace is the subspace generated by the basis {(1,0,0),(0,1,0)}. any vector in this subspace is of the form (a,b,0),
and T(a,b,0) = (2a+b,2b,0) = (2a+b)(1,0,0) + 2b(0,1,0).
so that's 6, you should be covered.