「統計解析のための線形代数」復習筆記17
正定，半正定 (正值，半正值)
對於任意的非零向量 \(\underline{x}(\neq\underline{0})\) ，如果2次型 \(\underline{x}^tA\underline{x}\) 始終滿足 \(\underline{x}^tA\underline{x} > 0\) 注意此處無等號。我們稱這個2次型爲正定(positive definite)，\(A\)爲正定矩陣(positive definite matrix)。另外，如果任意非零向量 \(\underline{x}(\neq\underline{0})\) 始終滿足2次型 \(\underline{x}^tA\underline{x} \geqslant 0\)， 這個2次型被叫做半正定(positive semi-definite)，\(A\)爲半正定矩陣(positive semi-definite matrix)。
\(x=\left( \begin{array}{} x_1\\ x_2\\ x_3 \end{array} \right), \ A=\left( \begin{array}{} 5 & 2 & 4\\ 2 & 2 & 3\\ 4 & 3 & 25 \end{array} \right)\)，2次型 \(\underline{x}^tA\underline{x}\) 是正定。因爲：
\(\underline{x}^tA\underline{x}=5x_1^2+2x_2^2+25x_3^2\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+4x_1x_2+8x_1x_3+6x_2x_3\\\;\;\;\;\;\;\;\;\;\:=(2x_1+x_2)^2+(x_2+3x_3)^2+(x_1+4x_3)^2\\ \because \underline{x}\neq\underline{0}=\left( \begin{array}{} 0\\ 0\\ 0 \end{array} \right)\\ \therefore \underline{x}^tA\underline{x}>0\)\(x=\left( \begin{array}{} x_1\\ x_2\\ x_3 \end{array} \right), \ A=\left( \begin{array}{} 5 & -6 & 3\\ -6 & 25 & 32\\ 3 & 32 & 73 \end{array} \right)\)，2次型 \(\underline{x}^tA\underline{x}\) 是半正定。因爲：
\(\underline{x}^tA\underline{x}=5x_1^2+25x_2^2+73x_3^2\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;-12x_1x_2+6x_2x_3+64x_1x_3\\ \;\;\;\;\;\;\;\;\;\:=(2x_1-3x_3)^2+(x_1+3x_3)^2+(4x_2+8x_3)^2\\\)
\(\because \underline{x}=\left( \begin{array}{c} 3\\ 2\\ -1 \end{array} \right)\) 時 \(\underline{x}^tA\underline{x}=0\)
\(\therefore \underline{x}^tA\underline{x} \geqslant0\)\(\underline{x}=\left( \begin{array}{c} x_1\\ x_2\\ \vdots\\ x_n \end{array} \right)(\neq\underline{0})\) 與單位矩陣 \(E_n\) 構成的2次型 \(\underline{x}^tE_n\underline{x}=\underline{x}^t\underline{x}=\sum\limits_{i=1}^nx_i^2>0\) 是爲正定。
\(\underline{x}=\left( \begin{array}{c} x_1\\ x_2\\ \vdots\\ x_n \end{array} \right), \underline{\frac{1}{n}}=\left( \begin{array}{c} \frac{1}{n}\\ \frac{1}{n}\\ \vdots \\ \frac{1}{n} \end{array} \right)\) 已知，\(\underline{x}\underline{1/n}=\sum\limits_{i=1}^nx_i\cdot \frac{1}{n}=\bar{x}\) (\(\underline{x}\) 的平均值)，包含了 \(n\) 個 \(\bar{x}\) 的橫向量： \((\bar{x}，\bar{x},\cdots,\bar{x})\) 展開以後成爲：
\((\bar{x}，\bar{x},\cdots,\bar{x})\\ \;\;\;\;\;\;\;\;\;\:=(x_1, x_2, \cdots, x_n)\left( \begin{array}{c} \frac{1}{n} & \frac{1}{n} & \cdots & \frac{1}{n}\\ \frac{1}{n} & \frac{1}{n} & \cdots & \frac{1}{n}\\ \vdots & \vdots & \ddots & \vdots \\ \frac{1}{n} & \frac{1}{n} & \cdots & \frac{1}{n} \end{array} \right)\\ \;\;\;\;\;\;\;\;\;\:=\underline{x}U\)
令 \(U\) 代表上面第二個等式中有半部分的矩陣。那麼將之從右往左乘以 \(\underline{x}\) 我們可以得到：
\(\underline{x}^tU\underline{x}=(\bar{x},\bar{x},\cdots,\bar{x})\underline{x}=(\bar{x},\bar{x},\cdots,\bar{x})\left( \begin{array}{c} x_1\\ x_2\\ \vdots\\ x_n \end{array} \right)\\ =\sum\limits_{i=1}^n\bar{x}x_i=\bar{x}\sum\limits_{i=1}^nx_i=n\bar{x}^2\)
利用上面的式子，我們可以得到，偏差平方和(sum of squared deviation, SS)：\(E_n\) 爲 \(n\) 次單位矩陣。
\(SS=\sum\limits_{i=1}^n(x_i-\bar{x})^2\\ \;\;\;\;\:=\sum\limits_{i=1}^n(x_i^2-2x_i\bar{x}+\bar{x}^2)\\ \;\;\;\;\:=\sum\limits_{i=1}^nx_i^2-2\bar{x}\sum\limits_{i=1}^nx_i+n\bar{x}^2\\ \;\;\;\;\:=\sum\limits_{i=1}^nx_i^2-2n\bar{x}^2+n\bar{x}^2\\ \;\;\;\;\:=\sum\limits_{i=1}^nx_i^2-n\bar{x}^2\\ \;\;\;\;\:=\sum\limits_{i=1}^nx_i^2-\underline{x}^tU\underline{x}\\ \;\;\;\;\:=\underline{x}^t\underline{x}-\underline{x}^tU\underline{x}\\ \;\;\;\;\:=\underline{x}^tE_n\underline{x}-\underline{x}^tU\underline{x}\\ \;\;\;\;\:=\underline{x}^t(E_n-U)\underline{x}\\ \because when \underline{x}=\left( \begin{array}{c} x_1\\ x_2\\ \vdots\\ x_n \end{array} \right)=\left( \begin{array}{c} \bar{x}\\ \bar{x}\\ \vdots\\ \bar{x} \end{array} \right), \& (\bar{x}\neq0), SS=0\\ \therefore \underline{x}^t(E_n-U)\underline{x}\;是半正定2次型。\)
雙一次型
對於 \(\underline{x}=\left( \begin{array}{c} x_1\\ x_2\\ \vdots\\ x_m \end{array} \right), A=\left( \begin{array}{c} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \vdots & \vdots\\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{array} \right), \underline{y}=\left( \begin{array}{c} y_1\\ y_2\\ \vdots\\ y_n \end{array} \right)\) 來說，\(\underline{x}^tA\underline{y}=\sum\limits_{i=1}^m\sum\limits_{j=1}^na_{ij}x_iy_j\) 既是 \(\underline{x}\) 的1次型，也是 \(\underline{y}\) 的1次型，所以又叫做 \(\underline{x}\underline{y}\) 的雙1次型(bilinear form)。雙1次型是一個標量(scalar)。
對於 \(\underline{x}=\left( \begin{array}{c} x_1\\ x_2\\ x_3 \end{array} \right), B=(b_{ij})=\left( \begin{array}{c} 1 & 0 & 0\\ 0 & 1 & 1\\ 1 & 0 & 1 \end{array} \right), \underline{y}=\left( \begin{array}{c} y_1\\ y_2\\ y_3 \end{array} \right)\)
\(\underline{x}^tB\underline{y}=\sum\limits_{i=1}^3\sum\limits_{j=1}^3b_{ij}x_iy_j=(x_1+x_3, x_2, x_2+x_3)\left( \begin{array}{c} y_1\\ y_2\\ y_3 \end{array} \right)\\ \;\;\;\;\:=x_1y_1+x_2y_2+x_2y_3+x_3y_1+x_3y_3\\ \;\;\;\;\:=x_1y_1+x_2(y_2+y_3)+x_3(y_1+y_3) \; \bf (\underline{x} 的1次型)\)
\(\;\;\;\;\;\:=y_1(x_1+x_3)+x_2y_2+(x_2+x_3)y_3 \;\bf (\underline{y} 的1次型)\)對於 \(\underline{l}=\left( \begin{array}{c} l_1\\ l_2\\ \end{array} \right), T=\left( \begin{array}{c} t_{11} & t_{12} & t_{13}\\ t_{21} & t_{22} & t_{23}\\ \end{array} \right), \underline{m}=\left( \begin{array}{c} m_1\\ m_2\\ m_3 \end{array} \right)\)
\(\underline{l}^tT\underline{m}=\sum\limits_{i=1}^2\sum\limits_{j=1}^3t_{ij}l_im_j\\ \;\;\;\;\;\;\;\;\;\;=l_1t_{11}m_1+l_1t_{12}m_2+l_1t_{13}m_3\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+l_2t_{21}m_1+l_2t_{22}m_2+l_3t_{23}m_3\\ \;\;\;\;\;\;\;\;\;\;=l_1(t_{11}m_1+t_{12}m_2+t_{13}m_3)\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+l_2(t_{21}m_1+t_{22}m_2+t_{23}m_3) \;\;(\underline{l} \textbf{的1次型})\\ \;\;\;\;\;\;\;\;\;\;=(t_{11}l_1+t_{21}l_2)m_1+(t_{12}l_1+t_{22}l_2)m_2\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+(t_{13}l_1+t_{23}l_2)m_3\;\;(\underline{m} \textbf{的1次型})\)
