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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
$$Q(\boldsymbol{x})=\boldsymbol{x}^TM(Q)\boldsymbol{x},$$
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_1x_1 \\ \vdots \\ r_nx_n\end{pmatrix} =r_1x_1^2+\cdots+r_nx_n^2. $
So the coefficients of $x_ix_j$ 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.
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