Conic section: What is the coordinate matrix of its bilinear form? Given is the conic section $x^2 + xy + y^2 + 2x +3y - 3 = 0$.
I need to find the coordinate matrix $M_\beta(s)$ of the bilinear form $s: \mathbb{R}^2 \times \mathbb{R}^2 -> \mathbb{R}$.
I read this article first: http://en.wikipedia.org/wiki/Matrix_representation_of_conic_sections
So I simply put
$A_Q = \begin{pmatrix} 1 & 0.5 & 1 \\ 0.5 & 1 & 1.5 \\ 1 & 1.5 & -3 \end{pmatrix}$ (Since A=B=C = 1, D=2, E=3 and F=-3).
So $A_Q$ is the coordinate matrix?

Keep in mind that a conic section is expressed through the matrix $A_Q$ as 
$$\mathbf{x}^T A\,\mathbf{x}=\begin{pmatrix}x & y & 1\end{pmatrix}
\begin{pmatrix}a & b & c \\ b & d & e\\ c & e &f \end{pmatrix}\!\begin{pmatrix}x \\ y \\ 1\end{pmatrix}=ax^2+2bxy+2cx+d y^2+2ey+f.$$
If we want the particular conic $x^2+xy+y^2+2x+3y-3=0$, we find by inspection that $(a,b,c,d,e,f)=(1,\frac12,1,1,\frac32,-3)$ which validates the answer given in the OP.
Moreover, observe that despite $\mathbf{x}\in\mathbb{R}^3$ it only depends on $(x,y)\in\mathbb{R}^2$. For this reason, the bilinear form $M\big((x_1,y_1),(x_2,y_2)\big)=\mathbf{x_1}^T A \,\mathbf{x}_2$ is computed using vectors in $\mathbb{R}^3$ but is really a mapping from $\mathbb{R}^2\times \mathbb{R}^2\to \mathbb{R}$ as was desired.
