知行合一，止于至善
千里之行，始于足下；合抱之木，生于毫末；九层之台，起于累土线性代数笔记
Transpose
(Ax)^T=x^TA^T\\ (AB)^T=B^TA^T\\ (AA^{-1})^T=(A^{-1})^T.A^T=I\rightarrow (A^T)^{-1}=(A^{-1})^T\\ \begin{bmatrix} a_1 & & \\ & \ddots \\ && a_n \end{bmatrix}^{-1} = \begin{bmatrix} \frac{1}{a_1} & & \\ & \ddots \\ && \frac{1}{a_n} \end{bmatrix}Determinants
det(AB)=det(A)\cdot det(B)
detA按行或按列展开a_{ij}\cdot(-1)^{i+j}\cdot\Delta_{ij}
projection
由
a^T(b-a\hat{x})=0
A^T(b-A\hat{x})=0\\
\hat{x}=(A^TA)^{-1}A^Tb\\
p=A\hat{x}=A(A^TA)^{-1}A^Tb\\
p=Pb \rightarrow P=A(A^TA)^{-1}A^T\\
eignvalue and eignvector
A\mathbf{x}=\lambda \mathbf{x} \\ (A-\lambda I)\mathbf{x}=0\\ \lambda_1 +\lambda_2 = \Sigma{a_{ii}}\\ \lambda_1 *\lambda_2 = detA\\S由A的特征向量构成的矩阵
AS=[\mathbf{s_1}\ \cdots\ \mathbf{s_n}]
\begin{bmatrix}
\lambda_1 & &\\
& \lambda_2 &\\
& & \lambda_3
\end{bmatrix}\\
AS = S\Lambda\\
A = S\Lambda S^{-1}\\
Symmetric
A=A^T
\lambda is real number
eignvector orthogonal
Q^TQ=I \\
A=Q\Lambda Q^T\\
least squares
求(0,6), (1,0) (2,0) 拟合直线
C+Dx=b \\
A=\begin{bmatrix}
1 & 0 \\
1 & 1 \\
1 & 2
\end{bmatrix}
x=\begin{bmatrix}
C\\
D\\
\end{bmatrix}
b=\begin{bmatrix}
6\\
0\\
0\\
\end{bmatrix}
x为b在A确定的平面上投影系数
SVD
(A^TA)v_i= \sigma _i^2 v_i 左乘v_i^T
v_i^TA^TAv_i = \sigma ^2v_i^Tv_i\ \ \ \ A^TA 为对称矩阵->特征向量为单位正交向量
(Av_i)^TAv_i = \sigma ^2\\
\begin{Vmatrix}Av_i\end{Vmatrix} = \sigma \\
AA^TAv_i = \sigma_i^2Av_i\\ 左乘矩阵A
u_i =Av_i/\sigma
那么
A[v_1 ... v_r] =[u_1 ... u_r]\begin{bmatrix}
\sigma_1 & &\\
& \sigma_2 &\\
& & \sigma_3
\end{bmatrix}\\
Av = u \Sigma
所以
A^TA=\mathbf{v}\Sigma^T\mathbf{u}^T.\mathbf{u}\Sigma \mathbf{v}^T
=\mathbf{v}\Sigma^2\mathbf{v}^T
\Sigma=A^TA的特征值\sqrt{ \lambda }，v=特征向量
u矩阵既可以通过AV/\Sigma 计算，亦可以和v一样通过AA^T计算，
A^TA和AA^T拥有相同的特征值
constant-coefficient differential equation
\frac{d\mathbf{u}}{dt}=A\mathbf{u}1、计算A的特征值和特征向量
2、方法1使用特征值和特征向量组成特殊解
\mathbf{u}=Ce^{\lambda_1 t}\mathbf{x_1} + De^{\lambda_2 t}\mathbf{x_2}
然后将初值u_{(0)}=u_0带入
Cx_1 + Dx_2 = u_0 确定系数
3、使用u=e^{At}\cdot u_0公式
e^{\Lambda t} = I+\Lambda t + \frac{1}{2}(\Lambda t)^2+ \frac{1}{6}(\Lambda t)^3 + \cdots + \frac{1}{n!}(\Lambda t)^n\\
= \begin{bmatrix}
e^{\lambda_1t} & & &\\
& e^{\lambda_2t} & &\\
& & \ddots &\\
& & & e^{\lambda_nt} &
\end{bmatrix} \\
e^{At} = I+At + \frac{1}{2}(At)^2+ \frac{1}{6}(At)^3 + \cdots + \frac{1}{n!}(At)^n\\
A^2=S\Lambda S^{-1}S\Lambda S^{-1} = S\Lambda^2 S^{-1}\\
A^n = S\Lambda^n S^{-1}\\
e^{At} = S(I+\Lambda t + \frac{1}{2}(\Lambda t)^2+ \frac{1}{6}(\Lambda t)^3 + \cdots + \frac{1}{n!}(\Lambda t)^n)S^{-1}\\
= Se^{\Lambda t}S^{-1}\\
u=e^{At}\cdot u_0= Se^{\Lambda t}S^{-1}u_0 = Se^{\Lambda t}C \\
4、二阶常系数微分方程
y^{\prime\prime} + y = 0 \\
\frac{dy}{dt} = y^{\prime}\\
y^{\prime\prime} = -y + 0y^{\prime}\\
\frac{d}{dt}
\begin{bmatrix}
y\\
y^{\prime}
\end{bmatrix} = \begin{bmatrix}
0 & 1 \\
-1& 0
\end{bmatrix}\begin{bmatrix}
y\\
y^{\prime}
\end{bmatrix}
Positeive Definite Matrices
symmmetric matric has one of these properties it has them all
1. all n pivots are positive
2 all n upper left determinants are positive
3 all n eigenvalues are positive
4 x^TAx is positeive except at x=0
5 A = R^TR , R with independent columns
ax^2+2bxy+cy^2 = 1
\begin{bmatrix}
x & y
\end{bmatrix}
\begin{bmatrix}
a & b\\
b & c\\
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix} \\
%%%%% 注释
\begin{bmatrix}
x & y
\end{bmatrix}
Q\Lambda Q^T
\begin{bmatrix}
x \\
y
\end{bmatrix}
=
\begin{bmatrix}
X & Y
\end{bmatrix}
\Lambda
\begin{bmatrix}
X \\
Y
\end{bmatrix}\rightarrow \\
\lambda_1X^2 + \lambda_2Y^2=1 \rightarrow\\
X = \frac {1}{\sqrt{\lambda_1}} \ \ Y = \frac {1}{\sqrt{\lambda_2}}
Complex Vectors and Matrices
\frac {1}{a+ib} = \frac {a-ib}{(a+ib)(a-ib)}\\
z=rcos\theta + irsin\theta = re^{i\theta} \\
z\cdot z^{\prime} = r\cdot r^{\prime}e^{i(\theta+\theta^{\prime})}\\
set \ \omega = e^{2\pi i/n} \rightarrow \\
(\omega)^n = (\omega^2)^n= (\omega^{n-1})^n=1
Hermitian Matrices ( A=A^H )
properties: 1)real eigenvalues 2) eigenvectors \perp
unitary Matrices(columns \perp )
U^HU=I
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