
dimension
let f: V> W. if the dim V = dim W, prove that f might not be an isomorphism.
my working:
in this case, if the dimV= rank f + nullity f= dim W,
it means that nullity f = 0.
for f to be an isomorphism, it means that it is bijective and linear.
so in this case, we need to prove that it is not linear.
let f(x) = (x)^2 + 1. then its proven.
can it be done like this?

To me it's not clear exactly what you are asking.
Are you asking: If is a linear map between vector spaces and with , show that need not be an isomorphism?
If so, just let be the linear map sending all elements of to the zero vector of (which won't be an isomorphism, whenever ).
And why should the equations imply that ? The map could easily have rank 1, and thus nonzero nullity (at least as far as ).
About nonlinear maps not being isomorphisms of vector spaces: Well, yeah, it is very much a true statement, but I really think that the map is implicitly assumed to be linear, so the author of the problem would probably like a linear example of a nonisomorphism.

i just took it that dim W = rank W = dim V and forgot the fact that the that is only true if f is an isomorphism.
in this case, from the example that you have given,
assuming that the dimV = n and all elements in V gets sent to the zero vector in W, does it mean that nullity is n and rank is 0 thus dimW= n?

In the example with the map sending everything from V (with dimV = n) to the zero vector, you are right that the nullity of the map is n, and that the rank is 0. You then write "thus dimW = n", which I don't quite get. Remember that you assumed to begin with that dimV = dimW, so if dimV=n, then dimW=n is immediately true as well.