「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)\(A\)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)\(A\)半正定矩陣(positive semi-definite matrix)

  1. \(x=\left( \begin{array}{} x_1\\ x_2\\ x_3 \end{array} \right), \ A=\left( \begin{array}{} 5 & 2 & 4\\ 2 & 2 & 3\\ 4 & 3 & 25 \end{array} \right)\),2次型 \(\underline{x}^tA\underline{x}\) 是正定。因爲:
    \(\underline{x}^tA\underline{x}=5x_1^2+2x_2^2+25x_3^2\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+4x_1x_2+8x_1x_3+6x_2x_3\\\;\;\;\;\;\;\;\;\;\:=(2x_1+x_2)^2+(x_2+3x_3)^2+(x_1+4x_3)^2\\ \because \underline{x}\neq\underline{0}=\left( \begin{array}{} 0\\ 0\\ 0 \end{array} \right)\\ \therefore \underline{x}^tA\underline{x}>0\)

  2. \(x=\left( \begin{array}{} x_1\\ x_2\\ x_3 \end{array} \right), \ A=\left( \begin{array}{} 5 & -6 & 3\\ -6 & 25 & 32\\ 3 & 32 & 73 \end{array} \right)\),2次型 \(\underline{x}^tA\underline{x}\) 是半正定。因爲:
    \(\underline{x}^tA\underline{x}=5x_1^2+25x_2^2+73x_3^2\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;-12x_1x_2+6x_2x_3+64x_1x_3\\ \;\;\;\;\;\;\;\;\;\:=(2x_1-3x_3)^2+(x_1+3x_3)^2+(4x_2+8x_3)^2\\\)
    \(\because \underline{x}=\left( \begin{array}{c} 3\\ 2\\ -1 \end{array} \right)\)\(\underline{x}^tA\underline{x}=0\)
    \(\therefore \underline{x}^tA\underline{x} \geqslant0\)

  3. \(\underline{x}=\left( \begin{array}{c} x_1\\ x_2\\ \vdots\\ x_n \end{array} \right)(\neq\underline{0})\) 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。

  4. \(\underline{x}=\left( \begin{array}{c} x_1\\ x_2\\ \vdots\\ x_n \end{array} \right), \underline{\frac{1}{n}}=\left( \begin{array}{c} \frac{1}{n}\\ \frac{1}{n}\\ \vdots \\ \frac{1}{n} \end{array} \right)\) 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 \(n\)\(\bar{x}\) 的橫向量: \((\bar{x},\bar{x},\cdots,\bar{x})\) 展開以後成爲:
    \((\bar{x},\bar{x},\cdots,\bar{x})\\ \;\;\;\;\;\;\;\;\;\:=(x_1, x_2, \cdots, x_n)\left( \begin{array}{c} \frac{1}{n} & \frac{1}{n} & \cdots & \frac{1}{n}\\ \frac{1}{n} & \frac{1}{n} & \cdots & \frac{1}{n}\\ \vdots & \vdots & \ddots & \vdots \\ \frac{1}{n} & \frac{1}{n} & \cdots & \frac{1}{n} \end{array} \right)\\ \;\;\;\;\;\;\;\;\;\:=\underline{x}U\)
    \(U\) 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)\(E_n\)\(n\) 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 \(\underline{x}\underline{y}\)雙1次型(bilinear form)。雙1次型是一個標量(scalar)。

  1. 對於 \(\underline{x}=\left( \begin{array}{c} x_1\\ x_2\\ x_3 \end{array} \right), B=(b_{ij})=\left( \begin{array}{c} 1 & 0 & 0\\ 0 & 1 & 1\\ 1 & 0 & 1 \end{array} \right), \underline{y}=\left( \begin{array}{c} y_1\\ y_2\\ y_3 \end{array} \right)\)
    \(\underline{x}^tB\underline{y}=\sum\limits_{i=1}^3\sum\limits_{j=1}^3b_{ij}x_iy_j=(x_1+x_3, x_2, x_2+x_3)\left( \begin{array}{c} y_1\\ y_2\\ y_3 \end{array} \right)\\ \;\;\;\;\:=x_1y_1+x_2y_2+x_2y_3+x_3y_1+x_3y_3\\ \;\;\;\;\:=x_1y_1+x_2(y_2+y_3)+x_3(y_1+y_3) \; \bf (\underline{x} 的1次型)\)
    \(\;\;\;\;\;\:=y_1(x_1+x_3)+x_2y_2+(x_2+x_3)y_3 \;\bf (\underline{y} 的1次型)\)

  2. 對於 \(\underline{l}=\left( \begin{array}{c} l_1\\ l_2\\ \end{array} \right), T=\left( \begin{array}{c} t_{11} & t_{12} & t_{13}\\ t_{21} & t_{22} & t_{23}\\ \end{array} \right), \underline{m}=\left( \begin{array}{c} m_1\\ m_2\\ m_3 \end{array} \right)\)
    \(\underline{l}^tT\underline{m}=\sum\limits_{i=1}^2\sum\limits_{j=1}^3t_{ij}l_im_j\\ \;\;\;\;\;\;\;\;\;\;=l_1t_{11}m_1+l_1t_{12}m_2+l_1t_{13}m_3\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+l_2t_{21}m_1+l_2t_{22}m_2+l_3t_{23}m_3\\ \;\;\;\;\;\;\;\;\;\;=l_1(t_{11}m_1+t_{12}m_2+t_{13}m_3)\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+l_2(t_{21}m_1+t_{22}m_2+t_{23}m_3) \;\;(\underline{l} \textbf{的1次型})\\ \;\;\;\;\;\;\;\;\;\;=(t_{11}l_1+t_{21}l_2)m_1+(t_{12}l_1+t_{22}l_2)m_2\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+(t_{13}l_1+t_{23}l_2)m_3\;\;(\underline{m} \textbf{的1次型})\)

Avatar
Chaochen Wang 王 超辰
Assistant Professor

All models are wrong, but some are useful.

comments powered by Disqus

Related