What is the direct sum of linear space and its dual space? For finite-dimensional linear space $L$ and its dual space $L^*$, take as basis of $L$ the set of vectors $\{e\}$, and as the basis of the dual space the dual basis $\{e^*\}$. Given $y\in L^*,x\in L$, denote $y(x)$ by the bracket $[x,y]$. I wish to show that $[x,y]$ is a bilinear form in $K=L \oplus L^* $, the direct sum of the linear space and its dual space.
I thought this was an interesting path to take, because the theory of bilinear forms is nicely developed, but I haven't found any information about it. I'm also not sure that it's 'true' (i.e. that it generalizes in a useful way that maintains the bilinear property), because I'm not sure what the direct sum of these spaces is, and how it would play into the bilinear form described.
It's clear that the linear space and its dual are isomorphic, so you could represent them by $n$-tuples, $x=(\xi_1,\xi_2,...,\xi_n)$, $y=(\eta_1,\eta_2,...,\eta_n)$, where $[x,y]$ is defined by
$$
[x,y] = \eta_1\xi_1+\eta_2\xi_2+...+\eta_n\xi_n
$$
But then we almost fall into the trap of thinking both $x$ and $y$ are in the same basis, which they are not. In the space $K$, we have to take $$x=\xi_1e_1+\xi_2e_2+...+\xi_ne_n+0e^*_1+0e^*_2+...+0e^*_n$$ $$y=0e_1+0e_2+...+0e_n+\eta_1e^*_1+\eta_2e^*_2+...+\eta_ne^*_n$$
I'm just not sure how to interpret general vectors $z$, and how to generalize the 'bilinear form' $[x,y]$ (quotes because it may not be a bilinear form for general vectors). Then there's the question of what $[x,x]$ or $[y,y]$ should mean. The whole of it comes down to the models underlying the vector spaces, so in order for this concept to be useful at all, the models need to generalize appropriately. I don't see any issue with the linear spaces themselves, apart from a partially defined mapping $[x,y]$ causing issues. I'm worried I'm going in the wrong direction though. Help!
--Edit--
By a bilinear form in a linear space $K$ over a field $\mathcal{F}$ is meant a mapping $A:K\times K\rightarrow\mathcal{F}$ such that $A(x,y)$ for $x,y\in K$ is linear in either argument when the other is fixed.
 A: I'm going to stick to the question you lay out in your first paragraph, which is pretty nice: If $L$ is a finite-dimensional vector space, we consider $K = L \oplus L^*$ and the form $(x, \alpha) \mapsto \alpha(x)$, where $x \in L$ and $\alpha \in L^*$. After the first paragraph you pick a basis and go off track a bit. Never pick a basis until you really need to.
So let's consider $q(x, \alpha) := \alpha(x)$. If $\lambda$ is a scalar (an element of the field these vector spaces are over), then we have
$$
q(\lambda \cdot (x, \alpha))
= q(\lambda x, \lambda \alpha)
= \lambda \alpha (\lambda x)
= \lambda^2 \alpha(x)
= \lambda^2 q(x, \alpha).
$$
Then what we have is a quadratic form on $L \oplus L^*$. There's an associated bilinear form $b$ on $L \oplus L^*$ defined by
\begin{align*}
b((x, \alpha), (y, \beta))
&= \frac12 q(x + y, \alpha + \beta) - \frac12 q(x, \alpha) - \frac12 q(y, \beta)
\\
&= \frac12 (\alpha + \beta)(x + y) - \frac12 \alpha(x) - \beta(y)
\\
&= \frac12 (\alpha(x) + \alpha(y) + \beta(x) + \beta(y)) - \frac12 \alpha(x) - \frac12 \beta(y)
\\
&= \frac12 (\alpha(y) + \beta(x)).
\end{align*}
This bilinear form satisfies $b((x, \alpha), (x, \alpha)) = q(x, \alpha)$, and is maybe what you were looking for.
There's a slightly different path we can take that also leads to the bilinear form $b$. To see it, note that if $b$ is a bilinear form on a vector space $V$, we can interpret $b$ as a linear map $\hat b : V \to V^*$. Concretely, this map is $\hat b(x) = (y \mapsto b(x, y))$. The two are related by $b(x,y) = \hat b(x)(y)$.
If we have an element $(x, \alpha)$ of $L \oplus L^*$, we can look for an element of its dual space $L^* \oplus L^{**}$. One that comes to mind is the element $(\alpha, \operatorname{ev}_x)$, where $\operatorname{ev}_x \in L^{**}$ is the evaluation map $\operatorname{ev}_x(\alpha) = \alpha(x)$. We then define
$$
\hat b : L \oplus L^* \to L^* \oplus L^{**},
\quad
\hat b(x, \alpha) = (\alpha, \operatorname{ev}_x).
$$
Contrary to $q$, this map is linear. The associated bilinear map is
$$
b((x, \alpha), (y, \beta))
= \hat b(x, \alpha)(y, \beta)
= (\alpha, \operatorname{ev}_x)(y, \beta)
= \alpha(y) + \beta(x),
$$
which is just the bilinear map we pulled out of $q$!
A: Question/remark: "I'm also not sure that it's 'true' (i.e. that it generalizes in a useful way that maintains the bilinear property), because I'm not sure what the direct sum of these spaces is, and how it would play into the bilinear form described."
Note: If $V$ is a vector space over a field $k$, there is a canonical isomorphism
$$\rho: V^*\otimes_k V \cong End_k(V)$$
defined by
$$\rho(\phi\otimes v)(u):=\phi(u)v.$$
There is a canonical map
$$tr: V^*\otimes_k V \rightarrow k$$
defined by
$$tr(\phi \otimes v):=\phi(v)$$
the "trace" of the endomorphism $\phi \otimes v \in End_k(V)$ and this "equals" your "product"
$$ [\phi, v]:=\phi(v) = tr(\phi \otimes v).$$
But as remarked in the comments: There is no natural $k$-linear map
$$V^* \oplus V \rightarrow End_k(V),$$
hence you cannot interpret $V^*\oplus V$ as a "sub-vector space of matrices" in $End_k(V)$.
