Determinant of a nondegenerate quadratic form Given a quadratic form $Q$ over a space $V$.  
To put it simply, we say $Q$ is nondegenerate if for every $x\in V$ there exists at least one $y\in V$ such that $x.y\neq 0$ (where '$.$' is a generalization of the inner product, and $x.y=\frac{1}{2}(Q(x+y)-Q(x)-Q(y)$). 
Equivalently, $Q$ is nondegenerate if $\det(A)\neq 0$, ($A$ is the matrix representation of $Q$).  
$\det(A)\neq 0$ means the kernel of $Q$ is ${0}$, hence $Q(x)=0$ iff $x=0$. But it contradicts the fact that we have isotropic elements ($0\neq x\in V$ such that $Q(x)=0$) in nondegenerate spaces.
what am I missing?  
Please try putting it as simple as you can.

Let's put the focus on the matrix $A$. We have 
$$Q(x) = x^T A x.$$
Suppose $Q(x) = 0$. This means $x^T (A x) = 0$, or equivalently, $x$ is orthogonal (under the standard inner product) to $Ax$. Note that this doesn't require that $Ax = 0$ - it only requires that $Ax$ be a vector which lives on the perpendicular subspace to $x$.
Example:
Let $$A = \begin{pmatrix}0 & 1 \\ 1 & 2\end{pmatrix}.$$
Then $$Q\begin{pmatrix}s \\ t \end{pmatrix} = \begin{pmatrix} s & t \end{pmatrix}\begin{pmatrix}0 & 1 \\ 1 & 2\end{pmatrix} \begin{pmatrix}s \\ t \end{pmatrix} = \begin{pmatrix} s & t \end{pmatrix} \begin{pmatrix} t \\ s + 2t\end{pmatrix} = 2(st+t^2).$$
Observe that $$Q\begin{pmatrix}1 \\ 0 \end{pmatrix} = 0,$$
however,
$$A \begin{pmatrix}1 \\ 0 \end{pmatrix} = \begin{pmatrix}0 \\ 1 \end{pmatrix} \neq 0.$$
