1) let $(v,w) \equiv v_1 w_2 + v_2 w_1$, then let check bilinearity
$(\alpha v, w) = \alpha v_1 w_2 + \alpha v_2 w_1 = \alpha (v_1 w_2 + v_2 w_1) = \alpha (v,w)$
$(v, \beta w) = v_1 \beta w_2 + v_2 \beta w_1 = \beta (v_1 w_2 + v_2 w_1) = \beta (v,w)$
$(v+q,w) = v_1 w_2 + q_1 w_2 + w_1 v_2 + w_1 q_2 = (v,w) + (q,w)$
and so on
2) symmetry $(v,w) = v_1 w_2 + v_2 w_1 = w_1 v_2 + w_2 v_1 = (w,v)$
3) positivity $(v,v) = v_1 v_2 + v_2 v_1 = 2 v_1 v_2$ which is not always $>0$. your definition failed here.
in contrast:
1) let $(v,w) \equiv 2 v_1 w_1 + (v_1 -v_2)(w_1-w_2) $,
positivity $(v,v) = 2 v_1 v_1 + (v_1 - v_2)(v_1-v_2) = 2 v_1^2 + (v_1-v_2)^2 > 0$ for any $v \ne 0 $
Let $V$ be a finite dimensional vector space over $\mathbb K$ (choose for example $\mathbb R$).
- On the dual space $V^{*}$: why?
The dual space $V^{*}$ is defined as
$$V^{*}:=\{ \varphi: V\rightarrow \mathbb K,~~ \varphi~ \text{linear}\};$$
in other words, the dual space is the space of all linear functionals, i.e. those maps from the original vector space $V$ to the ground field $\mathbb K$ which are linear.
Why is this new space interesting?
In a certain sense, introducing $V^{*}$ we used all information we had from the beginning to produce a new linear space. In fact, we started just with a pair $(V,\mathbb K)$ and we arrive at a new vector space $V^{*}$ whose elements "connect" $V$ to $\mathbb K$ respecting the non trivial structure on $V$, i.e. the linear space structure.
Moreover, if $V$ is finite dimensional, then $V^{*}$ is finite dimensional as well.
Let $\{e_i\}_{i=1,\dots,n}$ be a basis of $V$ ($n$ is the dimension of $V$). The dual set
$$\{\varphi_j\}_{j=1,\dots,n}$$
of elements $\varphi_j\in V^{*}$ s.t.
$$\varphi_j(e_i):=\delta_{ij}$$
is a basis of $V^{*}$. To prove it you can check any book of linear algebra.
Note that we used only the existence of the pair $(0,1)$, with $0:=\delta_{ij}$ for $i\neq j$ and $1:=\delta_{ij}$ for $i=j$ in the ground field $\mathbb K$ to introduce our basis $\{\varphi_j\}_{j=1,\dots,n}$.
Such pair $(0,1)$ always exists by definition of field: the $0$ is the zero element of the addition, while $1$ denotes the unit element of multiplicative composition.
- On $T:V\rightarrow V^{*}$
Let $T:V\rightarrow V^{*}$ be the linear map
$$T(v)(w):=\langle v,w\rangle, $$
denoting by $\langle \cdot,\cdot \rangle $ an inner product on $V$. We want to prove that $T$ is an isomorphism. We need to check that $T$ is injective and surjective.
On injectivity: as $T$ is linear, to prove injectivity is equivalent to show that
$$\operatorname{Ker}(T)=\{v\in V: T(v)=0 ~\text{in}~ V^{*}\}=\{0\}.$$
By definition, $T(v)=0$ in $ V^{*}$ (here "0" denotes the zero map in $V^{*}$!) if
$$\forall w\in V\Rightarrow T(v)(w)=\langle v,w\rangle=0~~(*)$$
in $\in\mathbb K$; $(*)$ holds if and only if $v=0$. In fact,
in $(*)$ we can choose $w=v$, arriving at
$$T(v)(v)=\langle v,v\rangle=0\Leftrightarrow v=0 $$
by definition of inner product (check it!). In other words we have proved that $T(v)=0\Leftrightarrow v=0$, i.e. $\operatorname{Ker}(T)=\{0\}$.
On surjectivity You need to prove that
$$\forall \varphi\in V^{*}~~ \exists v\in V : T(v)=\varphi$$
in $V^{*}$, i.e.
$$\forall w\in V~~ T(v)(w)=\varphi(w) $$
in $\mathbb K$. This last statement is equivalent to
$$\forall w\in V~~ \langle v,w\rangle=\varphi(w), $$
which can be easily proven using a base on $V$ and the dual basis on $V^{*}$, as above.
I leave it to you. I hope it helps.
Best Answer
The general method is the same as usual. Let $\langle \cdot, \cdot \rangle$ be the usual inner product and let $\langle x, y \rangle_A = \langle x, Ay \rangle$, where $A$ is a symmetric positive definite matrix.
Then given a matrix $B$, the adjoint $B^\square$ with respect to $\langle \cdot, \cdot \rangle_A$ is defined by $\langle B^\square x, y \rangle_A = \langle x, By \rangle_A$. Writing this out gives $\langle B^\square x, Ay \rangle = \langle x, ABy \rangle$, and letting $Ay = z$ gives $\langle B^\square x, z \rangle = \langle x, ABA^{-1}z \rangle = \langle A^{-1}B^* A x, z \rangle$.
It follows from is that $B^\square = A^{-1}B^* A$.
Using the notation in the question, we have $A^\square = {1 \over 3} \begin{bmatrix} 1 & -1 \\ 4 & 5\end{bmatrix}$.