How do I prove that this linear transformation is bijective? Let $V$ be a finite-dimensional inner product space (either real or complex) and $V^*=\mathrm{Hom}(V,k)$ be the dual space. Let $v\in V$. Define a mapping $L_v$ from $V$ to the ground field $k$ by $L_v(u)=\langle u,v\rangle$. Note that $L_v$ is linear.
Define
$$\begin{align}L:V&\to V^*\\v&\mapsto L_v\end{align}$$
Now, $L$ is linear if $V$ is a real inner product space and anti-linear if $V$ is a complex inner product space, i.e., $L(av+bw)=aL(v)+bL(w)$ in the real case, and $L(av+bw)=\bar aL(v)+\bar bL(w)$ in the complex case.
I would like to prove that $L$ is bijective.
Suppose $L(v)=L(w)$. Then, for all $u\in V$,
$$\begin{align}\langle u,v\rangle&=\langle u,w\rangle\\\langle u,v\rangle-\langle u,w\rangle&=0\\\langle u,v-w\rangle&=0\end{align}$$
The only vector that gives zero inner product with all vectors is the zero vector. Therefore, $v=w$, and $L$ is injective. It seems to me that surjectivity holds by construction. However, in the problem, there is a caution to "take care to account for the not-quite linearity of $L$ in the complex case". Am I missing something?
 A: It is difficult for me to tell what "It seems to me that surjectivity holds by construction." means. I think you are overlooking something here: abstractly an element $\psi \in V^*$ is simply a $k$-linear map $V \to k$, which a priori doesn't have much to do with the inner product which has apparently been chosen on $V$. I think the clearest way to see that your map $L$ is bijective given that it is injective is to appeal to the fact that $V$ and $V^*$ have the same dimension. (This is only true when $V$ is finite dimensional, but the map $v \mapsto L_v$ also fails to be surjective in the infinite-dimensional case!)
As for the problem's hint, the problem is just that you need to be clear what kind of thing $\langle \cdot, \cdot \rangle$ is. You haven't told us, but I'd assume that when $k = \mathbb{C}$ we are supposed to be given an inner product $\langle \cdot, \cdot \rangle$ on $V$, which in particular is complex antilinear in the second argument (this is called sesquilinear). Then $u \mapsto \langle u, v \rangle$ is a $\mathbb{C}$-linear map for each fixed $v \in V$, but the assignment $v \mapsto \langle \cdot, v \rangle$ is not itself $\mathbb{C}$-linear! Instead it is $\mathbb{C}$-antilinear, because the chosen sesquilinear form is too.
As it turns out this is not a problem when it comes to verifying that the map $L$ is bijective, but it is something you need to think about at first: for example, you have probably only shown that a linear map $L : V \to W$ between vector spaces is injective if and only if $L(v) = 0$ implies $v = 0$, but you haven't checked this for complex antilinear maps! Also, does our "dimensionality" argument work to establish surjectivity given that we aren't working with a linear map? Of course the answer is yes, but if you've never thought about it before, it's something you need to check.
A: I glossed over the proof of surjectivity. I must prove that there exists a $v\in V$ such that $L(v)=v^*$ for every linear transformation $v^*:V\to k$ in $V^*$.
Let $u_1,u_2,\ldots,u_n$ be an orthonormal basis for $V$. Since every $v^*\in V^*$ is a linear transformation from $V$ to $k$, every $v^*\in V^*$ can be represented by a mapping from $\{u_1,u_2,\ldots,u_n\}$ to $k$, i.e., we only need to specify $v^*(u_1),v^*(u_2),\ldots,v^*(u_n)\in k$. Now set $\langle u_1,v\rangle,\langle u_2,v\rangle,\ldots,\langle u_n,v\rangle$ to these elements, respectively. Since $v$ can be written as $\sum_{i=1}^n\lambda_iu_i$ for some $\lambda_1,\lambda_2,\ldots,\lambda_n\in k$, we have, for $j=1,2,\ldots,n$,
$$\begin{align}\langle u_j,v\rangle&=\langle u_j,\sum_{i=1}^n\lambda_iu_i\rangle\\&=\sum_{i=1}^n\bar{\lambda_i}\langle u_j,u_i\rangle\\&=\bar{\lambda_j}\end{align}$$
Now, $v^*(u_j)=\bar{\lambda_j}$ for $j=1,2,\ldots,n$. In the real case, they are the coordinates of $v$ with respect to $u_1,u_2,\ldots,u_n$. In the complex case, they are the complex conjugate of the coordinates of $v$ with respect to $u_1,u_2,\ldots,u_n$. In either case, there exists a $v\in V$ for every $v^*(u_1),v^*(u_2),\ldots,v^*(u_n)\in k$ we specify and thus for every $v^*\in V^*$.
