线代笔记

一、行列式

1.

$\left| \begin{matrix} x & b & b & \cdots & b \\ b & x & b & \cdots & b \\ b & b & x & \cdots & b \\ \vdots & \vdots & \vdots & & \vdots \\ b & b & b & \cdots & x \end{matrix}\right| = \left\lbrack {x + \left( {n - 1}\right) b}\right\rbrack {\left( x - b\right) }^{n - 1}$ .
若视 $x$ 为变量, $b$ 为常数,则行列式是 $x$ 的 $n$ 次多项式,其根是 ${x}_{1} = {x}_{2} = \cdots = {x}_{n - 1} = b,{x}_{n} = \left( {1 - n}\right) b$ .
当 $a$ 取 $\lambda - a$ 时, ${\left| \begin{matrix} \lambda - a & b & b & \cdots & b \\ b & \lambda - a & b & \cdots & b \\ b & b & \lambda - a & \cdots & b \\ \vdots & \vdots & \vdots & & \vdots \\ b & b & b & \cdots & \lambda - a \end{matrix}\right| }_{n \times n} = \left\lbrack {\lambda - a + \left( {n - 1}\right) b}\right\rbrack {\left( \lambda - a - b\right) }^{n - 1}$ .
若视 $\lambda$ 为变量, $a,b$ 为常数,则行列式是 $\lambda$ 的 $n$ 次多项式,其根是 ${\lambda }_{1} = {\lambda }_{2} = \cdots = {\lambda }_{n - 1} = a + b,{\lambda }_{n} = a - \left( {n - 1}\right) b$ .
后续可求特征值

2.

①若行列式为 “X” 型,则
a. 主副对角线上元素相同.

$$\left| \begin{matrix} a & & & & & b \\ & \ddots & & & \kern1mu \raisebox{1pt}{.} \kern2mu \raisebox{4pt}{.} \kern2mu \raisebox{7pt}{.} \kern1mu & \\ & & a & b & & \\ & & b & a & & \\ & \kern1mu \raisebox{1pt}{.} \kern2mu \raisebox{4pt}{.} \kern2mu \raisebox{7pt}{.} \kern1mu & & & \ddots & \\ b & & & & & a \end{matrix} \right|_{2n} = (a^2 - b^2)^n$$

b. 主副对角线上元素不同.

$$\left| \begin{matrix} {a}_{1} & & & & & {b}_{1} \\ & \ddots & & & \kern1mu \raisebox{1pt}{.} \kern2mu \raisebox{4pt}{.} \kern2mu \raisebox{7pt}{.} \kern1mu & \\ & & {a}_{k} & {b}_{k} & & \\ & & {b}_{k + 1} & {a}_{k + 1} & & \\ & \kern1mu \raisebox{1pt}{.} \kern2mu \raisebox{4pt}{.} \kern2mu \raisebox{7pt}{.} \kern1mu & & & \ddots & \\ {b}_{2k} & & & & & {a}_{2k} \end{matrix}\right|_{2k} = \mathop{\prod }\limits_{i = 1}^{k}\left( {a}_{i}{a}_{2k + 1 - i} - {b}_{i}{b}_{2k + 1 - i}\right) .$$

当 ${2k} = 4$ ,即 $k = 2$ 时, ${D}_{4} = \mathop{\prod }\limits_{i = 1}^{2}\left( {a}_{i}{a}_{5 - i} - {b}_{i}{b}_{5 - i}\right) = \left( {a}_{1}{a}_{4} - {b}_{1}{b}_{4}\right) \left( {a}_{2}{a}_{3} - {b}_{2}{b}_{3}\right)$ .
以上两个公式通常用于后续矩阵的正定问题、二次型问题,而非只用于计算行列式.
②如类对称(主对角线元素相同,左下、右上各元素相同)行列式,可通过递推法、归纳法计算.

$$\left| \begin{matrix} a & b & \cdots & b \\ c & a & \cdots & b \\ \vdots & \vdots & & \vdots \\ c & c & \cdots & a \end{matrix}\right| = \left\{ \begin{array}{l} \left\lbrack {a + \left( {n - 1}\right) b}\right\rbrack {\left( a - b\right) }^{n - 1},b = c, \\ \frac{b{\left( a - c\right) }^{n} - c{\left( a - b\right) }^{n}}{b - c},b \neq c. \end{array}\right.$$

③对于 $n$ 阶行列式,记 $\Delta = {k}^{2} - {4bc}$ ,有

$${\left| \begin{matrix} k & b & & \\ c & \ddots & \ddots & \\ & \ddots & \ddots & b \\ & & c & k \end{matrix}\right| }_{n} = \left\{ \begin{array}{l} \left( {n + 1}\right) \cdot {\left( \frac{k}{2}\right) }^{n},{k}^{2} = {4bc}, \\ \frac{\left( k + \sqrt{\bigtriangleup }\right) ^{n + 1} - {\left( k - \sqrt{\bigtriangleup }\right) }^{n + 1}}{2^{n + 1}\sqrt{\bigtriangleup }},{k}^{2} \neq {4bc} \end{array}\right.$$

3. 克拉默法则简单推导

如何理解克拉默法则？ - 知乎
就是通过构造一个这个样的矩阵：
$x_{i}=\operatorname* {d e t} \left( \begin{matrix} {1} & {0} & {\cdots} & {0} & {x_{1}} & {0} & {\cdots} & {\cdots} & {0} \\ {0} & {\ddots} & {\ddots} & {\vdots} & {x_{2}} & {\vdots} & {\vdots} & {\vdots} & {\vdots} \\ {\vdots} & {\ddots} & {1} & {0} & {\vdots} & {\vdots} & {\vdots} & {\vdots} & {\vdots} \\ {\vdots} & {\ddots} & {\ddots} & {1} & {\vdots} & {\vdots} & {\vdots} & {\vdots} & {\vdots} \\ {\vdots} & {\vdots} & {\vdots} & {0} & {x_{i}} & {0} & {\vdots} & {\vdots} & {\vdots} \\ {\vdots} & {\vdots} & {\vdots} & {\vdots} & {\vdots} & {1} & {\ddots} & {\vdots} & {\vdots} \\ {\vdots} & {\vdots} & {\vdots} & {\vdots} & {\vdots} & {0} & {1} & {\ddots} & {\vdots} \\ {\vdots} & {\vdots} & {\vdots} & {\vdots} & {x_{n-1}} & {\vdots} & {\ddots} & {\ddots} & {0} \\ {0} & {\cdots} & {\cdots} & {0} & {x_{n}} & {0} & {\cdots} & {0} & {1} \\ \end{matrix} \right)$

二、矩阵

1.

$$( A B C )^{\intercal}=C^{\intercal} B^{\intercal} A^{\intercal}$$

2. 二维下的旋转矩阵

$$M ( \theta)=\left[ \begin{array} {c c} {\operatorname{c o s} \theta\,} & {\,-\, s i n \theta} \\ \\ {s i n \theta\,} & {\, \operatorname{c o s} \theta} \\ \end{array} \right]=\operatorname{c o s} \theta \left[ \begin{array} {c c} {1} & {0} \\ {\, 0 \,} & {1} \\ \end{array} \right]+\operatorname{s i n} \theta \left[ \begin{array} {c c} {0} & {\,-1 \,} \\ {\, 1} & {\, 0 \,} \\ \end{array} \right]=\operatorname{e x p} \left( \theta \left[ \begin{array} {c c} {0} & {\,-1 \,} \\ {\, 1} & {\, 0 \,} \\ \end{array} \right] \right)$$

$$\left[ \begin{array}{cc} \cos \phi & -\sin \phi \\ \sin \phi & \cos \phi \end{array} \right] \begin{bmatrix} \rho\cos\theta \\ \rho\sin\theta \end{bmatrix} = \begin{bmatrix} \rho\cos(\theta+\phi) \\ \rho\sin(\theta+\phi) \end{bmatrix}$$