# $\operatorname{Hom}_k(k,V)$ is a vector space?

Is it true that a vector space is just the set of maps from the underlying field to the space itself. I.e. if $V$ is a vector space of the field $k$ then $$V\cong \operatorname{Hom}_k(k,V)$$ if so then this would make an intuitive understanding of the dual space $V^*$ somewhat trivial since $$V^{**}\cong V\cong \operatorname{Hom}_k(k,V)\implies V^*=\operatorname{Hom}_k(V,k)$$ If true, an explanation of why $V\cong \operatorname{Hom}_k(k,V)$ with a simple example or two would provide a lot of clarity for me since I could easily grasp the dual vector space idea from that point.

Edit: I actually had to read two proposed answers a couple times for the idea to sink in but I could only pick one answer.

• That isomorphism is indeed valid. I am not at all clear about what your "intuitive understanding is", per the implication you've written. I suppose you're saying that $[\operatorname{Hom}_k(k,V)]^* \cong \operatorname{Hom}_k(V,k)$ is "obviously true" for some reason. Is that so? – Omnomnomnom Aug 4 '16 at 18:20
• Yes, $[\operatorname{Hom}_k(k,V)]^* \cong \operatorname{Hom}_k(V,k)$ is how I arrived at the implication. – Bob Aug 4 '16 at 18:28
• Is there an intuitive justification you have for the isomorphism $[\operatorname{Hom}_k(k,V)]^* \cong \operatorname{Hom}_k(V,k)$? Note that this isomorphism (generally) fails to hold if $V$ is infinite dimensional. – Omnomnomnom Aug 4 '16 at 18:28
• What @Omnomnomnom says generally fails for infinite-dimensional spaces is $[\operatorname{Hom}_k(V,W)]^* \cong \operatorname{Hom}_k(W,V)$. – Henning Makholm Aug 4 '16 at 18:42
• More generally if $R$ is any unital ring and $M$ is an $R$-module, there is a natural isomorphism between $\hom_R(R,-)$ and the identity functor. (Also, @Omnomnomnom's last comment is incorrect.) – arctic tern Aug 4 '16 at 19:37

## 2 Answers

This works even for infinite-dimensional vector spaces (or for that matter for general modules over unital rings): The map $$f \in \operatorname{Hom}_k(k,V) \mapsto f(1) \in V$$ is always vector space isomorphism. You don't need duals for that.

This is clearly injective and a homomorphism; to see that it is surjective, note that $v\in V$ corresponds to the map $t\in k\mapsto t\cdot v\in V$.

Indeed, we have $\DeclareMathOperator{\Hom}{Hom} V \cong \Hom_k(k,V)$. To see that this is the case, note that we have the (natural) isomorphism $\Phi: V \to \Hom_k(k,V)$ given by $$[\Phi(v)](t) = t \cdot v$$ for $v \in V$ and $t \in k$.

In terms of matrices, this amounts to saying that a column-vector $v = (v_1,\dots,v_n) \in k^n$ can be naturally identified with the $n \times 1$ matrix $$\pmatrix{v_1\\ \vdots \\ v_n}$$ which denotes a linear map from $k^1$ to $k^n$.

• I'd appreciate feedback if I've said something inaccurate. – Omnomnomnom Aug 4 '16 at 22:26
• Makes sense to me as the OP. I marked it up in fact. Just wish I could mark both your answer and the other answer as correct. – Bob Aug 5 '16 at 1:30