# Isomorphism for finite dimensional Vector Spaces of same dimension over field F

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 ?

• $\rho$ could be the zero map – Hagen von Eitzen May 30 '17 at 22:02
• What about the map $\rho(x)=0$? There are many maps between $V$ and $V'$ that are not isomorphisms. – Rocket Man May 30 '17 at 22:02
• Yes, two vector spaces of the same dimension are isomorphic, but the map mentioned is not an isomorphism; is just linear map. Thus you can not deduce is bijective. – user 1987 May 30 '17 at 22:07
• @G.Sassatelli Sorry. Edited the same – SDG99 May 30 '17 at 22:17

Yes: take the null map. It's not an isomorphism, unless the common dimension is $0$.
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 $0$).
• Sorry for asking the stupid question. Yes I understand now. So as per the theorem, if $\rho$ is either injective or surjective, then it becomes a F-Vector Space Isomorphism. – SDG99 May 30 '17 at 22:20