I'm trying to prove the existence of the linear map $f^*$ in the following theorem about the adjoint:
Let $f: V \to W$ be a linear map, with $V$ finite-dimensional and $V,W$ inner product spaces over the same field. Then there is a unique linear map $f^*: W \to V$ such that $$\langle f(v),w \rangle = \langle v, f^*(w)\rangle$$ for all $v\in V, w \in W$.
Here's how far I got so far.
Try 1:
Fix $w \in W$. Because $V$ is finite-dimensional there exists a unique $x \in V$ for all linear forms $\psi_1 = \langle f(\cdot),w \rangle \in V^*$ --- namely we have the isomorphism $V \to V^*$. Since the linear form $\langle f(\cdot),w \rangle$ is uniquely determined by $w \in W$, define the linear map $f^*$ by $x = f^*(w)$. Let $\psi_2 = \langle \cdot,f^*(w) \rangle \in V^*$ and $\theta = \psi_1 - \psi_2 \in V^*$.
Now consider ker$(\theta)$. If ker$(\theta)$ = $V$ then our claim is proven --- namely $\psi_1 = \psi_2$.
Mistake: Firstly $\psi_1 \in W^*$, and not in $V^*$. And because $W \to W^*$ need not be an isomorphism the argument doens't follow.
EDIT: $\psi_1 \in V^*$ indeed. Namely $\psi_1: v \mapsto \langle f(v),w \rangle$ and $w \in W$ is fixed. However note that $\psi_1$ uses an inner product on $W$.
EDIT: Note that rk$(\theta) = 1$, so we have dim$\big($ker$(\theta)\big) = $ dim$(V) - 1$. Which fails my attempted proof. Namely take dim$(V) = 1$. So that, ker$(\theta)$ = $\{0\}$. And thus for all $0 \neq v \in V$ we have $\psi_1 \neq \psi_2$.
Try 2: the same as Try 1 but with $v = x$.
I'm assuming that the argument can be proven by a less cumbersome method. Also I think that my method is not correct, since I seem to get stuck.
Suggestions or an alternative method are welcome.
EDIT: Here's a summary of Fischer's proof.
Define the linear map $\lambda: W \to V^*$ by $$\lambda(w) = \langle \cdot, w \rangle_W \circ f.$$ Next define the isomorphism $\sigma: V \to V^*$ by $$\sigma(v_0) = \langle \cdot, v_0 \rangle_V.$$ Now, consider the composition $$\varphi: \sigma^{-1} \circ \lambda: W \to V.$$ We then have $$\langle v, \varphi(w) \rangle_V = \langle f(v), w \rangle_W$$ for all $v \in V, w \in W$. We can now define $f^*:= \varphi$. This proves the theorem for the existence part.
For uniqueness, assume that $\psi$ satisfies $\langle f(v),w \rangle_W = \langle v, \psi(w) \rangle_V$. Fixing $w_0 \in W$, we get $$\langle v,\psi(w_0) - f^*(w_0) \rangle_V = 0$$ for all $v \in V$. Choosing $v = \psi(w_0) - f^*(w_0)$ we get $\psi = f^*$, which proves the uniqueness part.