In the context of inner products and norms I need help with the following exercise :

First, some notation. In the vector space $$\mathbb R^2$$, we write $$\mathbf x = (x_1, x_2), \mathbf y = (y_1, y_2)$$ ect. for points $$\mathbf x, \mathbf y$$ in terms of a given base. For simplicity we assume that we are working over the field $$\mathbb R$$. Here's the exercise :

$$(1)$$ To which inner product corresponds the quadratic form

$$q(x_1, x_2) = 4x_1^2 - 6x_1x_2 + 5x_2^2$$

Find a vector perpendicular to $$(2, 3)$$ with respect to this inner product. Also find the angle between $$(-2, 3)$$ and $$(1, 2)$$.

Here's a simpler exercise which I was able to solve :

$$(2)$$ To which inner product corresponds the bilinear, symmetric function

$$f(\mathbf x, \mathbf y)= x_1y_1-x_2y_1-x_1y_2+4x_2y_2.$$

I used the following proposition : $$\langle x, y \rangle$$ is an inner product on $$\mathbb R^n$$ if and only if $$\langle x, y \rangle = x^TAy$$, where $$A$$ is a symmetric matrix whose eigenvalues are strictly positive.

So in this case the matrix is $$A=\begin{pmatrix}1&-1\\-1&4\end{pmatrix}$$ and the inner product is given by $$\langle x,y\rangle_f=x^TAy$$.

I wanted to use a similar argument for $$(1)$$ but I was not able to. One of the problems that I'm having is that in $$(2)$$ we are given a bilinear, symmetric function $$f(\mathbf x, \mathbf y)$$ which depends on $$\mathbf x$$ and $$\mathbf y$$ while in $$(1)$$ the quadratic form $$q$$ only depends on $$\mathbf x$$. Also, any help on the sub-questions would be appreciated.

Let $$f(\mathbf x) = \sum_{j=i}^n\sum_{i=1}^n a_{ij}x_ix_j = x^TAx$$ be a quadratic form. Then for $1\leq i \leq n$, $A_{ii}$ is given by the coefficient of $x_i^2$. For $i\neq j$, $A_{ij}=A_{ji}$ is given by one half of the coefficient of $x_ix_j$. For example, in your exercise, if $f(\mathbf x ) = 4x_1^2-6x_1x_2+5x_2^2$, then we have $$A = \begin{bmatrix} 4\,\,\,\,\,\,&-3 \\-3\,\,\,\,\,&5\end{bmatrix}$$ Then $\langle \mathbf x, \mathbf y\rangle = \mathbf x^T A\mathbf y$ is the inner product you're looking for.
To find a vector perpendicular to $(2,3)$, since the system of equations given by $(2,3)\cdot A\cdot\mathbf x =0$. Finally, the angle between two vectors $\mathbf u$ and $\mathbf v$ is given by $$\theta = \cos^{-1}\left(\frac{\langle \mathbf u, \mathbf v\rangle}{\sqrt{\langle \mathbf u, \mathbf u\rangle\langle \mathbf v, \mathbf v\rangle}} \right)$$