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A quadratic form is a homogenous polynomial $q(x_1, \dots , x_n) = \sum_{j=1}^n \sum_{i=1}^n x_i x_j a_{ij}$. Let $A := (a)_{ij}$. We define $q$ to be non-singular (or non-degenerate) if its associated bilinear form $B$ is non-singular where $ B(v,w) := \frac{1}{2} (q(v + w) - q(v) - q(w)) $

Why is singularity defined in terms of an associated bilinear form? Why not say $q$ is singular if and only if $A$ is?

Edit We may assume that $A$ is symmetric.

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    The matrix $A$ is not uniquely determined by $q$. You should require that $A$ is symmetric.2012-04-06

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Two points: (1)The expression $q(x_1,\cdots,x_n)=\sum_{i,j} a_{ij}x_ix_j$ is ambiguous in that it does not define $a_{ij}$ uniquely, as Andrea said. One of the matrices $A=(a_{ij})$ which determines $q$ in this sense (and the only symmetric one which does this) is the matrix of the associated bilinear form in the same basis. We can indeed characterize singular quadratic forms as those whose matrix (in this sense) is singular, in any (equivalently, in some) basis of the vector space. (2) Why do we define a "singular" (or "degenerate", which I'm more used to) quadratic form in terms of its associated bilinear form, rather than simply going to the associated matrix? Well, mainly because in Linear Algebra we'd rather have intrinsic definitions whenever they are at our disposal. That is, if some concept can be characterized in a coordinate-free way (without having to express the object at hand in terms of any basis of the vector space), we'll prefer these definitions.

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    So, fix a finite-dimensional vector space $V$. I'm denoting the elements of $V$ by $\bar v,$ and by $v$ the (column) n-tuple of the coordinates of $\bar v$ in a fixed basis. If you fix $V=\mathbb{R}^n$ and the canonical basis, this (subtle) difference dissapears.$A$quadratic form on $V$ is any map of the form $q(\bar v)=B(\bar v, \bar v)$ where $B$ is a symmetric bilinear form on $V$. "The" matrix of $q$ in the basis $\{\bar e_i\}$ is by definition "the" matrix of $B$ in that basis, its elements are exactly $a_{ij}=B(\bar e_i, \bar e_j)$, and it is always symmetric (because $B$ is symmetric)2012-04-07