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 ?

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    $\begingroup$ $\rho$ could be the zero map $\endgroup$ – Hagen von Eitzen May 30 '17 at 22:02
  • $\begingroup$ What about the map $\rho(x)=0$? There are many maps between $V$ and $V'$ that are not isomorphisms. $\endgroup$ – Rocket Man May 30 '17 at 22:02
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    $\begingroup$ 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. $\endgroup$ – user 1987 May 30 '17 at 22:07
  • $\begingroup$ @G.Sassatelli Sorry. Edited the same $\endgroup$ – SDG99 May 30 '17 at 22:17

Yes: take the null map. 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 $0$).

  • $\begingroup$ 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. $\endgroup$ – SDG99 May 30 '17 at 22:20
  • $\begingroup$ No question is stupid if it is to have a deeper understanding of a subject. To answer your question: yes. More precisely, it does not become – it is an isomorphism (provided the spaces have the same dimension). $\endgroup$ – Bernard May 30 '17 at 22:27

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