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

Proof

We consider the square matrix

M = {(α_{i j})}_{1 \leq i, j \leq n},

where

α_{i j} = {\begin{matrix} a_{i j} for i = j, \\ \frac{a_{i j}}{2} for i < j, \\ \frac{a_{j i}}{2} for i > j . \end{matrix}

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

(X_{1}, \dots, X_{n}) M (\begin{matrix} X_{1} \\ ⋮ \\ X_{n} \end{matrix}) .

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

B^{tr} M B

(describing the form with respect to the new basis, $B$ denotes the base change matrix) has diagonal form. Let $V_{1}, \dots, V_{n}$ 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