Quadratic polynomial/R/Change of variables/Pure form/Fact/Proof2

Proof

We consider the square matrix

${}M={\left(\alpha _{ij}\right)}_{1\leq i,j\leq n}\,,$

where

${}\alpha _{ij}={\begin{cases}a_{ij}{\text{ for }}i=j\,,\\{\frac {a_{ij}}{2}}{\text{ for }}i<j\,,\\{\frac {a_{ji}}{2}}{\text{ for }}i>j\,.\end{cases}}\,$

With this, the pure-quadratic term of the polynomial has the form

$\left(X_{1},\,\ldots ,\,X_{n}\right)M{\begin{pmatrix}X_{1}\\\vdots \\X_{n}\end{pmatrix}}.$

This equation holds upon inserting any element from ${}K$ for ${}X_{i}$, and also as an equation in ${}K[X_{1},\ldots ,X_{n}]$. By definition, this matrix ${}M$ is symmetric. Because of fact, there exists a orthonormal basis ${}v_{1},\ldots ,v_{n}$ of ${}\mathbb {R} ^{n}$ such that the new Gram matrix

${B^{\text{tr}}}MB$

(describing the form with respect to the new basis, ${}B$ denotes the base change matrix) has diagonal form. Let ${}V_{1},\ldots ,V_{n}$ be the new variables with respect to the new orthonormal system; that is, the ${}V_{i}$ describe, considered as functions, the linear forms to this new basis, that is, the dual basis. In the new variables, there are no mixed quadratic terms any more; the polynomial has now the form

${}F=\sum _{1\leq i\leq k}e_{i}V_{i}^{2}+\sum _{j=1}^{n}f_{j}V_{j}+g\,,$

with a certain ${}k$ between ${}1$ and ${}n$, and ${}e_{i}\neq 0$. The summands

$e_{i}V_{i}^{2}+f_{i}V_{i}$

can be brought, by completing the square and using new variables ${}U_{i}=V_{i}+h_{i}$, to the form

$e_{i}U_{i}^{2}+g_{i}.$

Besides the pure-quadratic term, either a constant or a linear polynomial remains. In the second case, we denote this linear form by ${}U_{k+1}$.

To fact