Section03_正定二次型

概念

• $f = \boldsymbol{X}^{\intercal}\boldsymbol{AX}$，若 $\forall\ \boldsymbol{X} \ne \boldsymbol{0}$ 有 $\boldsymbol{X}^{\intercal}\boldsymbol{AX} > 0$，称 $\boldsymbol{X}^{\intercal}\boldsymbol{AX}$ 为正定二次型，$\boldsymbol{A}$ 为正定矩阵

判别法

定义法

步骤

1. $\boldsymbol{A}^{\intercal}= \boldsymbol{A}$
2. $\forall\ \boldsymbol{X} \ne \boldsymbol{0}$，证 $\boldsymbol{X}^{\intercal}\boldsymbol{AX} > 0$

例题

1. $\boldsymbol{A},\boldsymbol{B}$ 正定，证 $\boldsymbol{A}+\boldsymbol{B}$ 正定 $$\begin{array}{ll} \because & \boldsymbol{A},\boldsymbol{B} \text{正定} \\ \therefore & \forall\ \boldsymbol{X}\ne \boldsymbol{0}, \boldsymbol{X}^{\intercal}\boldsymbol{AX} > 0, \boldsymbol{X}^{\intercal}\boldsymbol{B}\boldsymbol{X} > 0 \\ \therefore & \forall\ \boldsymbol{X} \ne 0, \boldsymbol{X}^{\intercal}(\boldsymbol{A}+ \boldsymbol{B})\boldsymbol{X}>0 \\ \therefore & \boldsymbol{A}+ \boldsymbol{B}\text{正定} \\ \end{array}$$
2. $\boldsymbol{A}_{m\times n}, r(\boldsymbol{A}) = n, \boldsymbol{B} = \boldsymbol{A}^{\intercal}\boldsymbol{A}$，证 $\boldsymbol{B}$ 正定 $$\begin{array}{ll} & \boldsymbol{B}^{\intercal} = \boldsymbol{A}^{\intercal}\boldsymbol{A} = \boldsymbol{B} \\\\ & \forall\ \boldsymbol{X} \ne \boldsymbol{0}, \boldsymbol{X}^{\intercal}\boldsymbol{B}\boldsymbol{X} = \boldsymbol{X}^{\intercal}\boldsymbol{A}^{\intercal} \boldsymbol{AX} = (\boldsymbol{AX})^{\intercal} \boldsymbol{AX} \\ \because & \boldsymbol{X}\ne 0, r(\boldsymbol{A}) = n \\ \therefore & \boldsymbol{AX} \ne \boldsymbol{0} \\ \therefore & (\boldsymbol{AX})^{\intercal}\boldsymbol{AX} = \vert \boldsymbol{AX} \vert^{2} > 0 \\ \therefore & \boldsymbol{B}\text{正定} \\ \end{array}$$

特征值法

方法

• Th $\boldsymbol{A}^{\intercal}\boldsymbol{A}$，则 $\boldsymbol{A}$ 正定 $\Leftrightarrow$ $\lambda_{i}> 0 \quad(1 \le i \le n)$

例题

1. $\boldsymbol{A}_{n\times n}$ 正定，证 $\boldsymbol{A}^{-1}$ 正定 $$\begin{array}{ll} \because & \boldsymbol{A}\text{正定} \\ \therefore & \lambda_{i} > 0, \boldsymbol{A}^{\intercal} = \boldsymbol{A}\\ \therefore & (\boldsymbol{A}^{-1})^{\intercal}= (\boldsymbol{A}^{\intercal})^{-1} = \boldsymbol{A}^{-1} \\ & \boldsymbol{A}^{-1}\text{的特征值} \frac{1}{\lambda_{i}} > 0\quad 1 \le i \le n \\ \therefore & \boldsymbol{A}^{-1}\text{正定} \\ \end{array}$$
2. $\boldsymbol{A}_{n\times n}$ 正定，证 $\det(\boldsymbol{A} + 2\boldsymbol{E})> 2^{n}$ $$\begin{array}{ll} \because & \boldsymbol{A}_{n\times n}\text{正定} \\ \therefore & \lambda_{i} > 0,\quad 1\le i\le n\\ \therefore & \boldsymbol{A} = 2 \boldsymbol{E}\text{的特征值}\lambda_{i} + 2 \\ \because & \det(\boldsymbol{A} + 2 \boldsymbol{E}) = \prod\limits_{i=1}^{n}(\lambda_{i} +2) \\ \therefore & \det(\boldsymbol{A} + 2\boldsymbol{E}) > 2^{n} \\ \end{array}$$

顺序主子式法

• $\boldsymbol{A}$ 正定 $\Leftrightarrow$ 所有顺序主子式大于零，即 $$\begin{array}{ll} & \text{对于} \boldsymbol{A} = \left( \begin{matrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{matrix} \right) \\ & \forall\ k \in [1, n],\ \left\vert \begin{matrix} a_{11} & \cdots & a_{1k} \\ \vdots & \ddots & \vdots \\ a_{k1} & \cdots & a_{kk} \end{matrix} \right\vert>0 \end{array}$$