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I have to find all matrices for which $\langle x,y\rangle_A = (x_1,x_2\ldots)A(y_1,y_2,\ldots)^T$ is a dot product $\forall x,y \in \mathbb{R}^2$ and $\forall A \in \mathbb{R}^{2\times2}$.

I already know that it has A to be symmertrical for the function to be symmetrical and for the positivity and the difitivety I got to the part where for some $\langle x, x\rangle_A$ with matrix A as $\begin{pmatrix} a &b\\ b&c \end{pmatrix}$ I got to $\langle x, x\rangle_A = ax_1^2+2bx_1x_2+cx_2^2$. but this is where i get stuck ( with the positivity) because what can i conclude from this?

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    There is no need for 2 $a$ in the matrix. The lower right entry can be $c$ too. – 2017-01-16
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    Good point but where to go from there? – 2017-01-16
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    First, I would think of $x_1=1$ and $x_2=0$ to get a first constraint. Then you can easily get a second one. – 2017-01-16
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    So a and c have to be positive – 2017-01-16
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    That's correct. Can you get something from the general case now, since you know the sign of $a$ and $c$ ? – 2017-01-16
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    Since $ax_1^2$ is always positive and $cx_2^2$ is always positive as well i would say$ax_1^2+cx_2^2 \ge 2bx_1x_2$? – 2017-01-16
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    Google "Positive definite matrix" – 2017-01-16

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Since $A$ has to be symmetric, write

$$ A = \begin{pmatrix} a & b \\ b & c \end{pmatrix}, \\ \left< (x,y), (x,y) \right>_A = ax^2 + 2bxy + cy^2.$$

We want $\left< (x,y), (x,y) \right> > 0$ for all $(x,y) \neq 0$. Since $\left< (1,0), (1,0) \right>_A = a$, we see that $a > 0$. Next, complete the square to get

$$ ax^2 + 2bxy + cy^2 = \left( \frac{ax + by}{\sqrt{a}} \right)^2 + \left( c - \frac{b^2}{a}\right) y^2.$$

From this expression, we see that if $c - \frac{b^2}{a} > 0$ (or, equivalently, $\det(A) = ac - b^2 > 0$), the expression will be positive if $(x,y) \neq (0,0)$. Finally, if $ac - b^2 \leq 0$ the any non-zero solution $(x,y)$ of $ax + by = 0$ will satisfy $\left< (x,y), (x,y) \right>_A \leq 0$.

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    Is there a rule which says that the expression $\det{(A)} = ac-b^2 >0$ is valid for any size of matrix or is this just a coincidence? – 2017-01-16
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    It's not a coincidence. Sylvester's criterion tells you that a matrix is positive definite if and only if all the principal minors (the determinants of the $k \times k$ square matrices obtained by looking only at the first $k$ rows and columns) are positive. For $2 \times 2$ matrix, the first minor is $a$ and the second is $\det(A)$. – 2017-01-16