from playLA.Matrix import Matrix def diagonalize(A): # 判断A是矩阵且是方阵 assert A.ndim == 2 assert A.shape[0] == A.shape[1] eigen_values, eigen_vectors = eig(A) # 矩阵对角化的P对应的就是特征向量构成的坐标转换矩阵 P = eigen_vectos # 通过矩阵的秩判断P是否可逆,不满秩则不可逆 ...
Let us see how to diagonalize a matrix using NumPy −Open Compiler import numpy as np # Define a matrix D = np.array([[2, 0, 0], [1, 3, 0], [4, 5, 6]]) # Compute the eigenvalues and eigenvectors eigenvalues, eigenvectors = np.linalg.eig(D) # Diagonal matrix of eigenvalues...
the eigenvectors are linearly independent and `a` can be diagonalized by a similarity transformation using `v`, i.e, ``inv(v) @ a @ v`` is diagonal. For non-Hermitian normal matrices the SciPy function `scipy.linalg.schur` is preferred because the matrix `v` is guaranteed to be unita...
although round-off error may obscure that fact. If the eigenvalues are all different, then theoretically the eigenvectors are linearly independent andacan be diagonalized by a similarity transformation usingv, i.e,inv(v) @ a @ vis diagonal. ...
Source File: 21-matrix_A_B.py From pyscf with Apache License 2.0 6 votes def diagonalize(a, b, nroots=4): a_aa, a_ab, a_bb = a b_aa, b_ab, b_bb = b nocc_a, nvir_a, nocc_b, nvir_b = a_ab.shape a_aa = a_aa.reshape((nocc_a*nvir_a,nocc_a*nvir_a)) a_...
Hi! Recently I'm reading your code in repo psi4/psi4numpy and know how to implement eom-ccsd method in python. In the process of reading the code, I'm confused about the code block where you make the sub-matrix of similarly transformed H...
The arrayvof eigenvectors may not be of maximum rank, that is, some of the columns may be linearly dependent, although round-off error may obscure that fact. If the eigenvalues are all different, then theoretically the eigenvectors are linearly independent andacan be diagonalized by a similarity...