Numpy 矩阵的特征值分解

>>> a
array([[1, 2, 3],
       [5, 8, 7],
       [1, 1, 1]])
       
>>> e_vals,e_vecs = np.linalg.eig(a)

>>> e_vals
array([ 10.254515  ,  -0.76464793,   0.51013292])

>>> e_vecs
array([[-0.24970571, -0.89654947,  0.54032982],
       [-0.95946634,  0.19306928, -0.73818337],
       [-0.13065753,  0.39865186,  0.40389232]])
# 注意上面矩阵的列向量才是特征向量，l2范数为 1

>>> [np.linalg.norm(e_vecs[:,i],ord=2) for i in [0,1,2]]
[0.99999999999999989, 0.99999999999999989, 1.0]

挑选出前 k 大的特征值与特征向量：

def topk(mat,k):
    e_vals,e_vecs = np.linalg.eig(mat)
    sorted_indices = np.argsort(e_vals)
    return e_vals[sorted_indices[:-k-1:-1]],e_vecs[:,sorted_indices[:-k-1:-1]]

a = np.array([[1,2,3],[5,8,7],[1,1,1]])
vals,vecs = topk(a,2)
print(vals)
print(vecs)

[ 10.254515     0.51013292]
[[-0.24970571  0.54032982]
 [-0.95946634 -0.73818337]
 [-0.13065753  0.40389232]]