以下のように二次形式を$x^Tx$から$y^TAy$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。
\[ \begin{array}{rcl} 左辺&=& \left[ \begin{array}{c} x & y \end{array} \right] \left[ \begin{array}{c} a_{11}x + a_{12}y\\ a_{21}x + a_{22}y\\ \end{array} \right] \\ &=& a_{11}x^2 + a_{12}xy + a_{21}xy + a_{22}y^2 \\ &=& a_{11}x^2 + (a_{12} + a_{21})xy + a_{22}y^2 \\ 右辺&=& (b_{11}x + b_{12}y)^2 + (b_{21}x + b_{22}y)^2 \\ &=& (b_{11}^2 + b_{21}^2)x^2 + 2(b_{11}b_{12} + b_{21}b_{22})xy + (b_{12}^2 + b_{22}^2)y^2 \end{array} \]$a_{12}=a_{21}$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。
\[ \left\{ \begin{array}{l} a_{11} = b_{11}^2 + b_{21}^2\\ a_{12} = a_{21} = b_{11}b_{12} + b_{21}b_{22}\\ a_{22} = b_{12}^2 + b_{22}^2 \end{array} \right. \]