the first one is easy..

i)

clearly, by taking an arbitrary element of and apply the mapping.

ii)

suppose , then there exist such that and .

now, since is an isomorphism. and since , so does . thus

iii)

Let and . Then such that

now, and just like the reasoning above, .

for the second one..

i)

let dim . take be a basis for . Let . then, there exists a such that , say ..

thus is a spanning set for

ii)

suppose (the zero vector for ).

then,

(the zero vector for ).. since be a basis for ,

..

this implies that is a basis for . Therefore dim .. QED