I recently came across the following statement in the book "A computational Introduction to Number Theory and Algebra" by Victor Shoup (Page:375)
Statement: Thus, two finite dimensional vector spaces are isomorphic if and only if they have the same dimension.
In the same page of the book, there is a theorem (Theorem:13.27) which states the following:
If $\rho: V \rightarrow V'$ is an F-Linear Map, and if $V$ and $V'$ are finite dimensional, with $dim_F(V) = dim_F(V')$, then we have: $ \rho $ is surjective iff $\rho$ is injective.
My question is, as per the statement, if two finite dimensional vector spaces have same dimension, then they are isomorphic, which should imply that the F-Linear Map $\rho$ defined in the theorem is a F-Vector Space isomorphism. This implies that $\rho$ is bijective. Why does the statement need to explicitly say $\rho$ is surjective if and only if $\rho$ is injective. As per the statement, $\rho$ is bound to be both surjective and injective since it the dimension of $V$ and $V'$ are same.
Is there a case where $\rho$ is not an isomorphism even if $dim_F(V) = dim(V')$ ?
Am I missing something ?
Yes: take the
nullmap. It's not an isomorphism, unless the common dimension is $0$.Saying that two vector spaces are isomorphic does not mean that all linear maps from one space to the other is an isomorphism.
Theorem 13.27 only gives a necessary and sufficient condition for a linear map between two vector spaces with the same dimension to be an isomorphism: it is enough to check either it is injective (so its kernel is $0$) or it is surjective (if you've heard of quotients spaces, this means its cokernel is $0$).