the first one is easy..
clearly, by taking an arbitrary element of and apply the mapping.
suppose , then there exist such that and .
now, since is an isomorphism. and since , so does . thus
Let and . Then such that
now, and just like the reasoning above, .
for the second one..
let dim . take be a basis for . Let . then, there exists a such that , say ..
thus is a spanning set for
suppose (the zero vector for ).
(the zero vector for ).. since be a basis for ,
this implies that is a basis for . Therefore dim .. QED