Let $V$ be a $\mathbb{K}$-vector space. Prove that $\operatorname{Hom}(\mathbb{K},V)$ is isomorphic to $V$.

Question

I was trying to solve this exercise:

Let $V$ be a $\mathbb{K}$-vector space. By definition we know $V^*=\operatorname{Hom}(V,\mathbb{K})$. Prove that $\operatorname{Hom}(\mathbb{K},V)$ is isomorphic to $V$.

My work:

Let $S=\operatorname{Hom}(V,\mathbb{K})$. Suppose $\dim(V)=n$. We just prove this $\dim(S)=n$.

In fact: $\dim(S)=\dim(V)\cdot \dim\mathbb{(K)}=n\cdot 1=n=\dim(V)$

Then $V$ is isomorphic to $V^*$

Is my argument correct?

Answer 1

Assuming you already know that $\dim(\textrm{Hom}(V,W)) = \dim(V)\dim(W)$, then yes, what you said here is correct. It will also be helpful to note the existence of an explicit dual basis: if $e_1,\ldots,e_n$ is a basis of $V$, then you can find $e_1^*,\ldots,e_n^*\in V^*$ such that $e_i^*(e_j) = \delta_{ij}$. If you write $v\in V$ uniquely as $\sum_{i=1}^n v_ie_i$, then $e_j^*(v) = v_j$. Knowing this outright will be helpful later.

Also, general vocabulary tip, they are called vector spaces, not vectorial spaces. And note that this proof is only in finite dimensional vector spaces, in infinite dimensions this is necessarily not true (the dual space is much larger)

Answer 2

The dual space has nothing to do with the statement you have to prove.

Consider instead the map $\varphi\colon\operatorname{Hom}(\mathbb{K},V)\to V$ defined, for a linear $f\colon\mathbb{K}\to V$, by $$ \varphi(f)=f(1) $$ You should be able to prove $\varphi$ is linear, injective and surjective, with no assumption at all on the dimension of $V$, which could well be infinite dimensional.