Find the symmetrical matrix $A$ so that $Q(\vec x) = \vec x^TA \vec x$ $Q(\vec x) = x_1^2+x_1x_2+x_2^2$
The matrix $A=\begin{bmatrix}1 & 0.5 \\ 0.5 & 1\end{bmatrix}$ seems to do the job. But what's the general procedure for finding a solution?
I can just think of setting it up like this for more clarity:
$\begin{bmatrix}x_1 & x_2\end{bmatrix}\begin{bmatrix}? & ?\\ ? & ?\end{bmatrix}\begin{bmatrix}x_1\\ x_2\end{bmatrix} =\begin{bmatrix}x_1^2+x_1x_2+x_2^2\end{bmatrix}$
But after that I'm lost, there surely must be some concepts I can apply.

$A$ is called the matrix associated with the quadratic form $Q$. The general procedure is rather simple: put the coefficients of $x_i^2$ in the diagonal $a_{ii}$ and divide the coefficient of $x_{ij}$ in $2$, writing it twice in $A$: once in $a_{ij}$ and once in $a_{ji}$.
In your example, the coefficient of $x_1^2$ is $1$ so $a_{11}=1$, the coefficient of $x_2^2$ is $1$ so $a_{22}=1$. The coefficient of $x_1x_2$ is $1$ so $a_{12}=a_{21}=\frac{1}{2}$, resulting in:
$$
A=\begin{pmatrix}
1 & 0.5 \\ 0.5 & 1
\end{pmatrix}
$$
Here is another example. Consider $Q(\underline{x})=x_1^2+2x_2^2+x_3^2+2x_1x_2+x_3x_2$. Then the matrix associated with $Q$ is:
$$
\begin{pmatrix}
1 & 1 & 0 \\
1 & 2 & 0.5\\
0 & 0.5 & 1
\end{pmatrix}
$$
For further information see here.
