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Homediagonal quadratic form

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# diagonal quadratic form

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 to a diagonal quadratic form. On the other hand, a quadratic form may be equivalent to more than one diagonal quadratic form.

## Mathematics Subject Classification

11E81*no label found*15A63

*no label found*11H55

*no label found*

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