Let $Q(\boldsymbol{x})\in k[x_{1},\ldots,x_{n}]$ be a quadratic form over a field $k$ ($\operatorname{char}(k)\neq 2$), where $\boldsymbol{x}$ is the column vector $(x_{1},\ldots,x_{n})^{T}$. We write $Q$ as

 $Q(\boldsymbol{x})=\boldsymbol{x}^{T}M(Q)\boldsymbol{x},$

where $M(Q)$ is the associated $n\times n$ symmetric matrix over $k$. We say that $Q$ is a diagonal quadratic form if $M(Q)$ is a diagonal matrix.

Let’s see what a diagonal quadratic form looks like. If $M=M(Q)$ is diagonal whose diagonal entry in cell $(i,i)$ is $r_{i}$, then

$Q(\boldsymbol{x})=\boldsymbol{x}^{T}\begin{pmatrix}r_{1}&\cdots&0\\ \vdots&\ddots&\vdots\\ 0&\cdots&r_{n}\end{pmatrix}\begin{pmatrix}x_{1}\\ \vdots\\ x_{n}\end{pmatrix}=\begin{pmatrix}x_{1}&\cdots&x_{n}\end{pmatrix}\begin{% pmatrix}r_{1}x_{1}\\ \vdots\\ r_{n}x_{n}\end{pmatrix}=r_{1}x_{1}^{2}+\cdots+r_{n}x_{n}^{2}.$

So the coefficients of $x_{i}x_{j}$ for $i\neq j$ are all $0$ in a diagonal quadratic form. A diagonal quadratic form is completely determined by the diagonal entries of $M(Q)$.

Remark. Every quadratic form is equivalent (http://planetmath.org/EquivalentQuadraticForms) to a diagonal quadratic form. On the other hand, a quadratic form may be to more than one diagonal quadratic form.

Title diagonal quadratic form DiagonalQuadraticForm 2013-03-22 15:42:05 2013-03-22 15:42:05 CWoo (3771) CWoo (3771) 12 CWoo (3771) Definition msc 11E81 msc 15A63 msc 11H55 canonical quadratic form DiagonalizationOfQuadraticForm