python与线性代数矩阵

1.标量相乘
每个元素与标量相乘
设A,B,CA,B,CA,B,C是相同维数的矩阵,rrr与s" role="presentation" style="position: relative;">sss为数,则由
a. A+BA+BA+B = B+AB+AB+A
b. (A+B)+C=A+(B+C)(A+B)+C=A+(B+C)(A+B)+C = A+(B+C)
c. (A+0)=A(A+0)=A(A+0) = A
d. r(A+B)=rA+rBr(A+B)=rA+rBr(A+B) = rA + rB
e. (r+s)A=rA+rB(r+s)A=rA+rB(r+s)A = rA + rB
f. r(sA)=(rs)Ar(sA)=(rs)Ar(sA) = (rs)A

2.矩阵相乘
若乘积ABABAB有定义,ABABAB的第iii行第j" role="presentation" style="position: relative;">jjj列的元素是AAA的第i" role="presentation" style="position: relative;">iii行与BBB的第j" role="presentation" style="position: relative;">jjj列对饮钙元素乘积之和.
只要注意矩阵左乘和右乘不相等即可.

a. A(BC)=(AB)CA(BC)=(AB)CA(BC) = (AB)C
b. A(B+C)=AB+ACA(B+C)=AB+ACA(B+C) = AB +AC
c. (B+C)A=BA+CA(B+C)A=BA+CA(B+C)A = BA +CA
d. r(AB)=(rA)B=A(rB)r(AB)=(rA)B=A(rB)r(AB) = (rA)B = A(rB)
e. ImA=A=AINImA=A=AINI_mA =A =AI_N

3.矩阵的转置(transpose)
ATATA^T行与列交换

a. (AT)T=A(AT)T=A(A^T)^T = A
b. (A+B)T=AT+BT(A+B)T=AT+BT(A+B)^T = A^T + B^T
c. (rA)=rAT(rA)=rAT(rA) = rA^T
d. (AB)T=BTAT(AB)T=BTAT(AB)^T = B^TA^T

4.矩阵的逆(inverse)
一个$n*n矩阵A是可逆的则

AA−1=IAA−1=IAA^{-1} = I

A−1A=IA−1A=IA^{-1}A = I
不可可逆矩阵也称为奇异矩阵(singular matrix),而可逆矩阵称为非奇异矩阵(nonsingular matrix)

a. 设A=[acbd][abcd] \left[ \begin{matrix}a & b \\c & d \\\end{matrix} \right] ,若ad!=bcad!=bcad!=bc,则A可逆,且A−1A−1A^{-1}=1ad−bc[a−c−bd]1ad−bc[a−b−cd]\frac{1}{ad-bc} \left[ \begin{matrix}a & -b \\-c & d \\\end{matrix} \right]

detA=ad−bcdetA=ad−bcdet A = ad - bc 称为A的行列式

b. 若A是可逆n∗nn∗nn*n矩阵,则对每一RnRnR^n中的b,方程Ax=bAx=bAx=b有唯一解x=A−1bx=A−1bx = A^{-1}b

a. (A−1)−1=A(A−1)−1=A(A^{-1})^{-1}=A
b. 若A和B都是n*n可逆矩阵,AB也可逆,且其逆是A和B的逆矩阵按相反顺序的乘积(AB)−1=B−1A−1(AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}
c. 若A可逆,则ATATA^T也可逆,且其逆是A−1A−1A^{-1}的转置,即(AT)−1=(A−1)T(AT)−1=(A−1)T(A^T)^{-1} = (A^{-1})^T

a.n*n矩阵A是可逆的,当且仅当A行等价于InInI_n,这时,把A变为InInI_n的一系列初等变换同时把InInI_n变成A−1A−1A^{-1}

可逆矩阵定理
a. A是可逆矩阵
b. A是n*n单位矩阵
c. A有n个主元位置
d.Ax=0Ax=0Ax =0仅有平凡解
e.A的个列线性无关
f.线性变换x−>Axx−>Axx -> Ax是一对一的
g. 对RnRnR^n中任意b,Ax=bAx=bAx = b至少有一个解
h. A的各列生成RnRnR^n
i. 线性变换x−>Axx−>Axx ->Ax把RnRnR^n映射到RnRnR^n上
j. 存在n*n矩阵C是的CA = I
k. 存在n*n矩阵D使得AD=I
l. ATATA^T是可逆矩阵

分块矩阵(partitioned matrices)
矩阵的因式分解(matrix factorizations)
RnRnR^n的子空间
维数与秩(dimension and rank)

python应用
在机器学习中KNN算法,要计算训练集A和测试集B中所有向量的欧式距离.使用二重嵌套循环效率会非常低,而用矩阵计算则会很快.

欧式距离OD = (x1−x2)2+(y1−y2)2−−−−−−−−−−−−−−−−−−√(x1−x2)2+(y1−y2)2\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}=x21+x22−2x1x2+y21+y22−2y1y2−−−−−−−−−−−−−−−−−−−−−−−−−−−√x12+x22−2x1x2+y12+y22−2y1y2\sqrt{x_1^2+x_2^2-2x_1x_2+y_1^2+y_2^2-2y_1y_2}

假设训练集A中有两个点分别是(a11,a12)和(a21,a22)(a11,a12)和(a21,a22)(a_{11} , a_{12})和(a_{21}, a_{22}),测试集B中有两个点分别是(b11,b12),(b21,b22),(b31,b32)(b11,b12),(b21,b22),(b31,b32)(b_{11} , b_{12}),(b_{21}, b_{22}),(b_{31}, b_{32})

第一种方法:二重嵌套循环

第二种方法:矩阵
首先,将A和B转换为矩阵
A=[a11a21a12a22][a11a12a21a22] \left[ \begin{matrix}a_{11} & a_{12} \\a_{21} & a_{22} \\\end{matrix} \right] B=⎡⎣⎢b11b21b31b12b22b32⎤⎦⎥[b11b12b21b22b31b32] \left[ \begin{matrix}b_{11} & b_{12} \\b_{21} & b_{22} \\b_{31} & b_{32}\end{matrix} \right]
最后得到的结果应该是一个2*3的矩阵,第i行第j列元素对应的A矩阵中第i个点和b矩阵中第j个点

欧式距离可以写成
欧式距离OD = (x1−x2)2+(y1−y2)2−−−−−−−−−−−−−−−−−−√(x1−x2)2+(y1−y2)2\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}=x21+x22+y21+y22−2x1x2−2y1y2−−−−−−−−−−−−−−−−−−−−−−−−−−−√x12+x22+y12+y22−2x1x2−2y1y2\sqrt{x_1^2+x_2^2+y_1^2+y_2^2-2x_1x_2-2y_1y_2}

1.首先计算x21+x22x12+x22x_1^2+x_2^2
AsAsA_{s}=[a211+a212a221+a222][a112+a122a212+a222] \left[ \begin{matrix}a_{11}^2 +a_{12}^2 \\a_{21}^2 +a_{22}^2 \\\end{matrix} \right]

同理计算y21+y22y12+y22y_1^2+y_2^2
⎡⎣⎢⎢b211+b212b221+b222b231+b232⎤⎦⎥⎥[b112+b122b212+b222b312+b322] \left[ \begin{matrix}b_{11}^2 + b_{12}^2 \\b_{21}^2 + b_{22}^2 \\b_{31}^2 + b_{32}^2\end{matrix} \right]

2.再求x21+x22+y21+y22x12+x22+y12+y22x_1^2+x_2^2+y_1^2+y_2^2
这里牵涉到矩阵的广播(broadcast)
AsbAsbA_{sb} = [a211+a212a221+a222a211+a212a221+a222a211+a212a221+a222][a112+a122a112+a122a112+a122a212+a222a212+a222a212+a222] \left[ \begin{matrix}a_{11}^2 +a_{12}^2 & a_{11}^2 +a_{12}^2 & a_{11}^2 +a_{12}^2\\a_{21}^2 +a_{22}^2 & a_{21}^2 +a_{22}^2 & a_{21}^2 +a_{22}^2 \\\end{matrix} \right]

这里使用b的转置来求
先求B的转置
[b211+b212b221+b222b231+b232][b112+b122b212+b222b312+b322] \left[ \begin{matrix}b_{11}^2 + b_{12}^2 &b_{21}^2 + b_{22}^2 &b_{31}^2 + b_{32}^2\end{matrix} \right]
在求BsBsB_s转置之后的广播

BsbBsbB_{sb} = [b211+b212b211+b212b221+b222b221+b222b231+b232b231+b232][b112+b122b212+b222b312+b322b112+b122b212+b222b312+b322] \left[ \begin{matrix}b_{11}^2 + b_{12}^2 &b_{21}^2 + b_{22}^2 &b_{31}^2 + b_{32}^2 \\b_{11}^2 + b_{12}^2 &b_{21}^2 + b_{22}^2 &b_{31}^2 + b_{32}^2\end{matrix} \right]

然后将AsbBsbAsbBsbA_{sb}B_{sb}相加
Asb+BsbAsb+BsbA_{sb}+B_{sb} = [a211+a212+b211+b212a221+a222+b211+b212a211+a212+b221+b222a221+a222+b221+b222a211+a212+b231+b232a221+a222+b231+b232][a112+a122+b112+b122a112+a122+b212+b222a112+a122+b312+b322a212+a222+b112+b122a212+a222+b212+b222a212+a222+b312+b322] \left[ \begin{matrix}a_{11}^2 +a_{12}^2+b_{11}^2 + b_{12}^2 &a_{11}^2 +a_{12}^2+b_{21}^2 + b_{22}^2 &a_{11}^2 +a_{12}^2+b_{31}^2 + b_{32}^2 \\a_{21}^2 +a_{22}^2+b_{11}^2 + b_{12}^2 &a_{21}^2 +a_{22}^2+b_{21}^2 + b_{22}^2 &a_{21}^2 +a_{22}^2+b_{31}^2 + b_{32}^2\end{matrix} \right]

3.求2x1x2+2y1y22x1x2+2y1y22x_1x_2+2y_1y_2
ABTABTAB^T=[b211+b212b221+b222b231+b232][b112+b122b212+b222b312+b322] \left[ \begin{matrix}b_{11}^2 + b_{12}^2 &b_{21}^2 + b_{22}^2 &b_{31}^2 + b_{32}^2\end{matrix} \right]

参考文献:
http://blog.csdn.net/Jiajing_Guo/article/details/62217564

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