$\displaystyle \ T : V \rightarrow W$ 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) $\displaystyle \le$ dim(W) ( don't know how to show the inequality other way around)