I have a question regarding the quadratic forms and their associated matrices. For some reason, google keeps telling me that this matrix is symmetric. The definition I am using is that a quadratic form on an R-module $E$ is a function $E\rightarrow R$ so that $q(rx)=r^2q(x)$ and the function $B(x,y)=q(x+y)-q(x)-q(y)$ is a symmetric bilinear form on $E$. Now, we can represent $B(x,y)$ in the form of a symmetric matrix $A$, such that $B(x,y)=x^TAy$. I'm a little confused as to why this means that we can represent $q$ by a matrix, but I assume that you can "pull out" the top right and bottom left part of $A$ to create a matrix C and its transpose, and after re-arranging and collecting terms we get $(x+y)^TC(x+y)-x^TCx-y^TCy$ so then $q$ must be given by $C$? But why does the internet say that $C$ must be symmetric? If we don't have a field certainly it mustn't be symmetric. If we just consider $\mathbb{Z}^2$, and the quadratic form $q([x,y])=x^2+y^2-xy$ the matrix $C$ is then the matrix with the first row $[1, -1]$ and the second row $[0,1]$. My next question is, if I know $C$ is positive definite can I conclude $A=C+C^T$ is positive definite? The ultimate goal is to somehow prove that the associated bilinear form of a positive definite quadratic form is nondegenerate when $R=\mathbb{Z}$ and $E=\mathbb{Z}^n$. Hope someone can help me understand where my confusion lies, and how to attack the problem.
$\underline{\textbf{EDIT:}} \\$ I forgot to include my attempt at the problem. So if we associate to $q$ the matrix $C$, we know $x^TCx> 0$ for all $x\in \mathbb{Z}^n$ not equal to $0$. Now, $A=C+C^T$. Then we want to show, that there does not exist an $x$ not equal to zero such that $B(x,y)=x^TAy=0$ for all $y$. The claim is to take any $x\neq 0$, and consider $y=x$. Then $x^TCx=x^TC^Tx>0$, so that $x^TAx>0$. I think this proves it. I hope this is right, I've confused myself so much thinking about the stuff above that I'm at the point that I might be making up magic math haha.