「統計解析のための線形代数」復習筆記16

二次型(形式)

對於 \(\underline{x}=\left( \begin{array}{c} x_{1}\\ x_{2}\\ \vdots\\ x_{n} \end{array} \right), A=\left( \begin{array}{c} a_{11}& a_{12} & \cdots & a_{1n}\\ a_{21}& a_{22} & \cdots & a_{2n}\\ \vdots& \vdots & \vdots & \vdots\\ a_{n1}& a_{n2} & \cdots & a_{nn} \end{array} \right)\) 那麼:
\(\underline{x}^tA\underline{x}=\sum\limits_{i=1}^n\sum\limits_{j=1}^na_{ij}x_ix_j\\ \;\;\;\;\;\;\;\;\:\:=\sum\limits_{i=1}^na_{ii}x_i^2+\mathop{\sum\limits^n\sum\limits^n}_{i \neq j}a_{ij}x_ix_j\\ \;\;\;\;\;\;\;\;\:\:=\sum\limits_{i=1}^na_{ii}x_i^2+\mathop{\sum\limits^n\sum\limits^n}_{i\ <\ j}(a_{ij}+a_{ji})x_ix_j\)
被稱爲 \(\underline{x}\) 的同次2次式。又被叫做關於 \(x_1,x_2,\cdots,x_n\)2次型(quadratic form)。特別的,當 \(A\) 爲對稱矩陣時的2次型:\(\underline{x}^tA\underline{x}=\sum\limits_{i=1}^na_{ii}x_i^2+2\mathop{\sum\limits^n\sum\limits^n}_{i\ <\ j}a_{ij}x_ix_j\) 在多元變量分析中十分重要。

  1. \(x=\left( \begin{array}{} x_1\\ x_2 \end{array} \right),\ A=\left( \begin{array}{} a_{11} & a_{12}\\ a_{12} & a_{22} \end{array} \right)\), 那麼:
    \(\underline{x}^tA\underline{x}=a_{11}x_1^2+a_{22}x_2^2+2a_{12}x_1x_2\)

  2. \(x=\left( \begin{array}{} x_1\\ x_2\\ x_3 \end{array} \right),\ A=\left( \begin{array}{} a_{11} & a_{12} & a_{13}\\ a_{12} & a_{22} & a_{23}\\ a_{13} & a_{23} & a_{33} \end{array} \right)\),那麼:
    \(\underline{x}^tA\underline{x}=a_{11}x_1^2+a_{22}x_2^2+a_{33}x_3^2+2a_{12}x_1x_2\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+2a_{13}x_1x_3+2a_{23}x_2x_3\)

  3. \(x=\left( \begin{array}{} x_1\\ x_2\\ x_3 \end{array} \right), \ A=\left( \begin{array}{} 3 & 2 & 4\\ 6 & 5 & 1\\ -2 & 5 & 8 \end{array} \right)非對稱矩陣\),那麼:
    \(\underline{x}^tA\underline{x}=3x_1^2+5x_2^2+8x_3^2\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+2x_1x_2+4x_1x_3+6x_2x_1\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+x_2x_3-2x_3x_1+5x_3x_2\\ \;\;\;\;\;\;\;\;\;\:=3x_1^2+5x_2^2+8x_3^2\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+(2+6)x_1x_2+(4-2)x_1x_3\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+(5+1)x_2x_3\\ \;\;\;\;\;\;\;\;\;\:=3x_1^2+5x_2^2+8x_3^2\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+8x_1x_2+2x_1x_3+6z_2x_3\)

  4. 有式子 \(3x_1^2-5x_2^2+7x_3^2+8x_1x_2+4x_1x_3-12x_2x_3\) 如果要將它改寫成 \(\underline{x}^tA\underline{x}\) (\(A\)是對稱矩陣) 的話,試求 \(A=\left( \begin{array}{} a_{11} & a_{12} & a_{13}\\ a_{22} & a_{22} & a_{23}\\ a_{33} & a_{32} & a_{33} \end{array} \right)=\left( \begin{array}{} a_{11} & a_{12} & a_{13}\\ a_{12} & a_{22} & a_{23}\\ a_{13} & a_{23} & a_{33} \end{array} \right)\) 的各個成分。
    \(A\) 的對角成分:\(a_{11},a_{22},a_{33}\) 分別是 \(x_1^2,x_2^2,x_3^2\) 的系數: \(3,-5,7\)。非對角成分 \(a_{12}(=a_{21})\) 等於 \(x_1x_2\) 系數的一半:\(4\)\(a_{13}(=a_{31})\) 等於 \(x_1x_3\) 系數的一半:\(2\), \(a_23(=a_{32})\) 等於 \(x_2x_3\) 系數的一半:\(-6\)
    \(\therefore A=\left( \begin{array}{} 3 & 4 & 2\\ 4 & -5 & -6\\ 2 & -6 & 7 \end{array} \right)\)

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