内積から二次形式へ

以下のように二次形式を$x^{T}x$から$y^{T}Ay$へ変換するとき、行列の中身はどうなるのか？確認してみた。

$$\left[\begin{array}{cc}x& y\end{array}\right]\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{cc}{b}_{11}x+b_{12}y& b_{21}x+b_{22}y\end{array}\right]\left[\begin{array}{c}{b}_{11}x+b_{12}y\\ b_{21}x+b_{22}y\end{array}\right]$$

左辺と右辺をそれぞれ展開して共通項を見てみる。

$$\begin{array}{rcl}\mathrm{左}\mathrm{辺}& =& \left[\begin{array}{cc}x& y\end{array}\right]\left[\begin{array}{c}{a}_{11}x+a_{12}y\\ a_{21}x+a_{22}y\end{array}\right]\\ & =& a_{11}{x}^2+a_{12}xy+a_{21}xy+a_{22}{y}^2\\ & =& a_{11}{x}^2+(a_{12}+a_{21})xy+a_{22}{y}^2\\ \mathrm{右}\mathrm{辺}& =& (b_{11}x+b_{12}y{)}^2+(b_{21}x+b_{22}y{)}^2\\ & =& (b_{11}^2+b_{21}^2)x^2+2(b_{11}{b}_{12}+b_{21}{b}_{22})xy+(b_{12}^2+b_{22}^2)y^2\end{array}$$

$a_{12}=a_{21}$に注意して両辺の共通項を取り出すと以下のようになる。

$$\{\begin{array}{l}{a}_{11}=b_{11}^2+b_{21}^2\\ a_{12}=a_{21}=b_{11}{b}_{12}+b_{21}{b}_{22}\\ a_{22}=b_{12}^2+b_{22}^2\end{array}$$