there was also the assumption dim(V) = dim (W) so that makes it easier.
is a linear transformation, V and W vector spaces
1) T is injective
2) T is surjective
3) Rank(T) = Dim(V)
3) ---> 1) nullity(T) + Rank(T) = Dim(V) , so 0 = nullity(T) iff T is injective
2) ----->3) T surjective implies that Dim(W) = Rank(T) (i don't know how to proceed from here)
1)-----> 2) T injective iff nullity(T) = 0 , which implies Rank(T) = Dim (V), so the number of linearly indep vectors to form a basis for the image = Rank(T), so Rank(T) dim(W) ( don't know how to show the inequality other way around)