# Matrix of a quadratic form?

What exactly is the matrix of a quadratic form? I have seen this notation occuring in a few papers (e.g. Siegel's unreadable German papers), with particular reference to the trace of a quadratic form. I'm at a loss as to what this means, and as a bonus question in passing I'd be interested if the trace of a quadratic form was interesting for an "obvious" reason (echoes of character theory, maybe?)

For example what is the matrix of the quadratic form $x^2+y^2+z^2$? Or $x^2+xy+y^2$?

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It is a matrix $A$ so that $q(x) = \langle x, A x \rangle$, where $q$ is your quadratic form. $A$ can (and is usually) taken to be symmetric (assuming reals here). In your first example, $A$ is the $3\times 3$ identity matrix, for the second it is a $2\times 2$ matrix with ones on the diagonal and $\frac{1}{2}$ on the off diagonal. – copper.hat Nov 3 '12 at 23:57

The matrix of a quadratic form $Q$ is the symmetric matrix $A$ such that

$$Q(\vec{x}) = \vec{x}^T A \vec{x}$$

For example,

$$x^2 + xy + y^2 = \left(\begin{matrix}x & y \end{matrix}\right) \left(\begin{matrix}1 & \frac{1}{2} \\ \frac{1}{2} & 1 \end{matrix}\right) \left(\begin{matrix}x \\ y \end{matrix}\right)$$

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For many authors, the matrix is actually double the ones given by martini and Hurkyl, so that the matrix is, quite simply, the Hessian matrix of the function, meaning the second partials. This is the more common behavior when discussing "even" lattices, as in most of NEBE. The importance of doubling everything is to arrange that all inner products of vectors in the lattice be integers. The most important example here is the root lattices of Lie algebras. So, with that in mind see PAGE $E_8$ where "the" matrix is given as

4 -2 0 0 0 0 0 1
-2 2 -1 0 0 0 0 0
0 -1 2 -1 0 0 0 0
0 0 -1 2 -1 0 0 0
0 0 0 -1 2 -1 0 0
0 0 0 0 -1 2 -1 0
0 0 0 0 0 -1 2 0
1 0 0 0 0 0 0 2


under the word GRAM.

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A quadratic form $q \colon V \to K$ on a vector space $V$ arises as "diagonal" of a symmetric bilinear form $\beta\colon V \times V \to K$ via $q(v) = \beta(v,v)$. Given $V$ has a basis $(v_i)$, we can associate a matrix $B = (b_{ij})$ with $\beta$, where $b_{ij} = \beta(v_i, v_j)$. For $v = \sum \lambda_i v_i$, $w = \sum\mu_i v_i \in V$ we have then $\beta(v,w) = \sum_{i,j} \lambda_i b_{ij} \mu_j = (\lambda_i)_i^T B (\mu_j)_j,$ so for $q$ we have $q(v,v) = (\lambda_i)^T B (\lambda_i)$ For your examples: $\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{pmatrix}\text{ and } \begin{pmatrix} 1 & \frac 12 \\ \frac 12 & 1 \end{pmatrix}.$

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Wow, you're fast... – copper.hat Nov 3 '12 at 23:57
And 24 seconds faster than me. :( – Hurkyl Nov 3 '12 at 23:58