Without invoking Riesz's representation theorem, why do Hodge duals exist? Working through Peter Szekeres's Course in Modern Mathematical Physics and I'm running into some confusion in its development of the Hodge dual. In particular, I'm not clear on what justification the book is using for its existence.
Choose a basis and let $E$ be its volume element. Let $A\in\Lambda^p(V)$ let the map $f_A:\Lambda^{n-p}(V)\to\mathbb R$ be defined by $A\land B=f_A(B)E$. It's easy to check that $f_A$ is a linear functional. Szekeres claims on page 223:

...as the inner product $(\cdot,\cdot)$ on $\Lambda^{n-p}(V)$ is non-singular there exists a unique $(n-p)$-vector $\ast A$ such that $f_A(B)=(\ast A,B)$...

Ignoring the "non-singular" part, this claim is clearly immediately true by the Riesz representation theorem. However, this bit of algebra/analysis isn't introduced in the text until way later. What I'm interested in is what justification is being used here.
What the book defines as being non-singular is the following property of inner products on page 127:

If $u\cdot v=0$ for all $v\in V$ then $u=0$.

Am I missing something obvious? I don't see the chain of reasoning here.
 A: In the finite-dimensional case, the Riesz theorem is pretty much trivial for non-degenerate bilinear forms, because it follows almost immediately from definition. 
Let $W$ be a finite-dimensional vector space over $\Bbb{R}$ (in your case, it's the $p^{th}$ exterior power), and $g$ is a non-degenerate bilinear form on $W$, i.e $g:W \times W \to \Bbb{R}$ is bilinear, and
\begin{align}
\text{for all $x\in W$, if  for all $y \in W$, $g(x,y) = 0$ then $x=0$.}
\end{align}
Note that another way to state the definition of non-degeneracy is as follows. We shall define a linear map $g^{\flat}:W \to W^*$ by the rule:
\begin{align}
g^{\flat}(x) &:= g(x, \cdot) \in W^*
\end{align}
In other words, for any $x \in W$,  $g^{\flat}(x) \in W^*$ is that element such that for all $y\in W$, we have
\begin{align}
(g^{\flat}(x))(y):= g(x,y).
\end{align}
Then, the statement of non-degeneracy is that
\begin{align}
g^{\flat}(x) = 0 \implies x = 0.
\end{align}
It's a basic linear algebra theorem that this is equivalent to saying $g^{\flat}:W \to W^*$ is injective. Now, since $W$ is finite-dimensional, $W^*$ has the same dimension, so $g^{\flat}$ is actually an isomorphism. The inverse map is typically denoted $g^{\sharp} := (g^{\flat})^{-1}:W^* \to W$.
Now, note that for all $f\in W^*$, and all $x \in W$, we have
\begin{align}
f(x) &= [(g^{\flat} \circ g^{\sharp})(f)](x) \tag{since $g^{\flat} \circ g^{\sharp} = \text{id}_{W^*}$} \\
&= [g^{\flat}(g^{\sharp}(f))](x) \\
&:= g\left( g^{\sharp}(f), x \right) \tag{$\ddot{\smile}$}
\end{align}

In your case, $f_A \in W^* = \left( \Lambda^{n-p}(V)\right)^*$. Since $(\cdot, \cdot) =g$ is non-degenerate, it provides us with an isomorphism $g^{\sharp}:\left( \Lambda^{n-p}(V)\right)^* \to \Lambda^{n-p}(V)$. Hence, the "vector" associated with the "covector" $f_A$ is
\begin{align}
\star A &:= g^{\sharp}(f_A) \equiv \bigg((\cdot, \cdot) \bigg)^{\sharp}(f_A) \in \Lambda^{n-p}(V).
\end{align}
Hence, non-degeneracy of $(\cdot, \cdot)$ implies the existence of a unique $\star A \in W = \Lambda^{n-p}(V)$, and if you compare with formula $(\ddot{\smile})$, you'll see that for all $B$, we have
\begin{align}
f_A(B) &= (\star A, B).
\end{align}

By the way, the maps $g^{\flat}$ (read as "$g$-flat") and $g^{\sharp}$ (read as "$g$-sharp") are called the musical isomorphisms.
