Learning about dual vector spaces I began thinking about how the product of a row vector and a column vector acts like a dot product and that furthermore, the row vector is almost acting like a functional that takes in the column vector and outputs a real number. This made me think of the following way to prove that all vector spaces are isomorphic to their double dual:
First, recall that the set of all column vectors with $n$ entries forms a $n$-dimensional vector space. Furthermore, every vector space of dimension $n$ is isomorphic to the vector space of column vectors with $n$ entries, $\mathscr C_n$. This can be shown through a simple change in basis. Therefore to show that any finite-dimensional vector space $V \cong V^{**}$, it suffices to show that $\mathscr C_n \cong \mathscr C_n^{**}$.
Consider the map $\phi : \mathscr C_n \rightarrow \mathscr C_n^*$ defined such that $$\forall \mathbf v,\mathbf x \in \mathscr C_n : \phi(\mathbf v)(\mathbf x) = \mathbf v^T\mathbf x$$ If we can show that this map is (1) one-to-one, (2) onto, and (3) linear, then it follows that the image of any basis in $\mathscr C_n$ under $\phi$ is a basis of $\mathscr C^*$ and so $\phi$ is an isomorphism.
Denote the vector space of all row vectors with n entries as $\mathscr R_n$. Then
$$\forall \mathbf f \in \mathscr R_n : \exists! \mathbf v \in \mathscr C_n : \mathbf f = \mathbf v^T \implies \forall \omega \in \mathscr C_n^* : \exists! \mathbf v \in \mathscr C_n : \phi(\mathbf v) = \omega$$ So $\phi$ is (1) one-to-one and (2) onto.
Let $\mathbb F$ denote the underlying field for $\mathscr C_n, \mathscr R_n,$ and consequently $\mathscr C^*$. Now, note that $\forall \mathbf v,\mathbf w,\mathbf x \in \mathscr C_n, c \in \mathbb F:$
$$\phi(c\mathbf v)(\mathbf x) = (c\mathbf v)^T\mathbf x = c(\mathbf v^T\mathbf x) = c\phi(\mathbf v)(\mathbf x)\\\phi(\mathbf v+\mathbf w)(\mathbf x)=(\mathbf v+\mathbf w)^T\mathbf x=(\mathbf v^T+\mathbf w^T)\mathbf x=\mathbf v^T\mathbf x+\mathbf w^T\mathbf x=\phi(\mathbf v)(\mathbf x)+\phi(\mathbf w)(\mathbf x)$$ And so $\phi$ is (3) linear. Therefore $\phi$ is an isomorphism and so $\mathscr C_n \cong \mathscr C_n^* \implies \mathscr C_n^* \cong \mathscr C_n^{**}$, which implies that any finite-dimensional vector space $V$ is isomorphic to $V^{**}$.
I know of the much simpler canonical isomorphism between $V$ and $V^{**}$, I just want to test my knowledge and see if this proof is also valid. Any feedback would be appreciated!
