矩阵identity（numpy仿真）

∑ixTiAxi=tr(A∑ixixTi)=tr(AXXT)

import numpy as np

def main():
    X = np.random.randn(10, 3)
    A = np.random.randn(3, 3)
    sum1 = 0
    for i in range(X.shape[0]):
        sum1 += np.dot(X[i, :], np.dot(A, X[i, :].T))
    sum2 = np.trace(np.dot(A, np.dot(X.T, X)))
    print(sum1)
                        # 10.158513956
    print(sum2)
                        # 10.158513956
if __name__ == '__main__':
    main()

特征值分解

令 A 是一个 N×N 的方阵， 且有 N 个线性无关的特征向量 qi(i=1,2,…,N) 。这样 A 可被分解为：

A=QΛQ−1⇓AQ=QΛ

>>> import numpy as np
>>> A = np.random.randn(3, 3)

>>> Lambda, Q = np.linalg.eig(A)
                    # 其中Lambda是一维向量，由特征值构成

>>> np.dot(Q, np.dot(np.diag(Lambda), np.linalg.inv(Q)))
array([[-0.15446862, -1.57279859, -0.28165496], [-0.99437763, -0.54065794, 0.75029032], [-0.49977911, 1.78752911, -1.17139559]])
>>> A
array([[-0.15446862, -1.57279859, -0.28165496], [-0.99437763, -0.54065794, 0.75029032], [-0.49977911, 1.78752911, -1.17139559]])

>>> [np.linalg.norm(Q[:, i], 2) for i in range(Q.shape[1])]
                        # 每一个特征向量的二范数，
[0.99999999999999989, 1.0, 1.0]
                        # 可见，numpy提供的特征分解已为我们做了特征向量的归一化

实对称矩阵不同的特征值对应的特征向量彼此正交（证明见 矩阵理论拾遗）。为了使用numpy线性代数工具箱进行测试，我们首先构造一个对称矩阵：

>>> import numpy as np
>>> A = np.random.randn(3, 3)
>>> A = np.triu(A)
>>> A += (A.T-np.diag(A.diagonal()))

>>> Lambda, Q = np.linalg.eig(A)
>>> np.dot(Q.T, Q)
[[ 1.00000000e+00 -3.88578059e-16 1.11022302e-16] [ -3.88578059e-16 1.00000000e+00 1.11022302e-16] [ 1.11022302e-16 1.11022302e-16 1.00000000e+00]]

实对称矩阵又可被分解为：

A=QΛQT

也即 QT=Q−1⇒QQT=I ， Q 为 正交矩阵（orthogonal matrix）；

import numpy as np

def main():
    X = np.random.randn(10, 3)  
    N = X.shape[0]
    C = np.dot(X.T, X)/N
    Lambda, Q = np.linalg.eig(C)

    print(np.dot(Q, Q.T))
    print(np.dot(Q.T, Q))

if __name__ == '__main__':
    main()

QTQ=I ，正交矩阵的行列式比为1或-1，

1=det(I)=det(QTQ)=det(QT)det(Q)=(detQ)2⇓det(Q)=±1

矩阵分块

C=1N∑xixTi=1NXXT

import numpy as np

def main():
    X = np.random.randn(10, 3)  
    C = 0
    N = X.shape[0]
    for i in range(N):
        C += np.dot(X[i][:, np.newaxis], X[i][np.nexaxis, :])
    C /= N
    C2 = np.dot(X.T, X)/N
    print(C==C2)

if __name__ == '__main__':
    main()

C=UΛUT=∑αλαuαuTα

二次型

xTAx=∑i,jxixjAij

import numpy as np

def main():
    x = np.array([1, 2, 3]) 
    A = np.random.randn(3, 3)

    print(np.dot(x, np.dot(A, x)))

    s = 0
    for i in range(A.shape[0]):
        for j in range(A.shape[1]):
            s += A[i, j]*x[i]*x[j]  
    print(s)

if __name__ == '__main__':
    main()

全1矩阵左乘一个矩阵 ≠ 右乘一个矩阵

[1,1,11]⋅[a11,a21,a12a22]=[11][1,1]⋅[a11,a21,a12a22]

列和在重复；

[a11,a21,a12a22]⋅[1,1,11]=[a11,a21,a12a22]⋅[11]⋅[1,1]

行和在重复；