That still only gives a 1D image space: see reasoning below.
We know that
e1, e2∈R2 are linearly independent since they are the standard basis vectors. So it is possible to construct a 2-dimensional image space with
f(e1), f(e2) as the basis vectors.
So all we need to do for part (c) for the 2-dimensional case is to choose a,b,c,d,e make sure that
f(e1)=(a,b,0), f(e2)=(c,d,e)∈R3 are linearly independent.
As analogues, for the 1-dimensional case we have two options:
1. Choose a,b,c,d,e such that either one of
f(e1)=(a,b,0), f(e2)=(c,d,e) is the zero vector and the other one isn't.
2. Choose a,b,c,d,e such that
f(e1)=(a,b,0), f(e2)=(c,d,e) are scalar multiples of each other and neither is the zero vector.
So your suggestion b=d=1, a=c=e=0 would make both
f(e1)=(0,1,0), f(e2)=(0,1,0) - they are the same vector so are scalar multiples of each other and hence linearly dependent so you can't have both in the basis.
For the 0-dimensional case, we only have one option: that
f(e1)=(a,b,0), f(e2)=(c,d,e) are both the zero vector, since we don't want any vectors in the basis.
This is the fastest way I can think of for doing part (c).