Will $f(v,w) = w^T A v$ define an inner product where $v,w \in \Bbb{R}^2$ and $A = \begin{bmatrix} 1 & 0 \\ 0 & -1\end{bmatrix}$?
If $\langle x,y\rangle = x^T Ay$ then for that to be an inner product $A$ must be symmetric, positive definite that is eigenvalues must be positive, but in my case eigenvalues of $A$ are $1,-1$, so is this not an inner product, also how $f(v,w) = f(w,v)$?
On
For an arbitrary $n \times n$ matrix $A$ we can consider the map $f(v,w) = v^T Aw$. Then $f$ is always bilinear, and
$f$ is an inner product iff $A$ is symmetric and positive-definite.
If $A$ is symmetric then $$f(v,w) = v^T Aw = (v^T Aw)^T = w^T A^T v = w^T Av = f(w,v)$$ so $f$ is symmetric. Conversely if $f$ is symmetric, then for any $1 \le i,j \le n$ $$A_{ij} = e_i^T Ae_j = f(e_i,e_j) = f(e_j,e_j) = A_{ji}$$ where $\{e_1,\ldots,e_n\}$ is the standard basis of $\mathbb{R}^n$, so $A$ is symmetric.
Finally, $f(v,v) > 0$ for all $v \ne 0$ iff $A$ is positive-definite, directly from the definition.
This doesn't quite define an inner product, but the only thing that fails as far as I can tell is $f(v,v)\geq0$ and $f(v,v)=0\iff v=0$. Consider $v=( 1,1)^T$ or $v=(1,2)^T$ for those facts.
As far as the symmetricity you asked about, that does actually hold. One easy way to check that is to choose variables for the coordinates of $v$ and $w$ and verify that the multiplication works as desired. If $v=( a,b)^T$ and $w=(c,d)^T$, then $f(v,w)=f(w,v)=ac-bd$.