Question

我是ARPACK的新手，我下载了类似以下的脚本

import time
import numpy as np
from scipy.linalg import eigh
from scipy.sparse.linalg import eigs
np.set_printoptions(suppress=True)

n=30
rstart=0
rend=n

A=np.zeros(shape=(n,n))

# first row
if rstart == 0:
    A[0, :2] = [2, -1]
    rstart += 1
# last row
if rend == n:
    A[n-1, -2:] = [-1, 2]
    rend -= 1
# other rows
for i in range(rstart, rend):
    A[i, i-1:i+2] = [-1, 2, -1]

A[0,8]=30

start_time = time.time()

evals_large, evecs_large = eigs(A, 10, sigma=3.6766133, which='LM')
print evals_large

end_time=time.time()-start_time
print(" Elapsed time: %12f seconds " % end_time)

它解决了一个非常简单的特征值问题（矩阵A不对称，我将A[0,8]设置为30）。根据ARPACK结果，最接近3.6766133（设置中为sigma=3.6766133）的3个特征值为

[ 3.68402411+0.j  3.82005897+0.j  3.51120293+0.j]

然后我转到MATLAB，并解决相同的特征值问题，结果是

4.144524409923138 + 0.000000000000000i 
3.642801014184622 + 0.497479798520641i
3.642801014184622 - 0.497479798520641i
2.372392770347609 + 0.762183281789166i
2.372392770347609 - 0.762183281789166i
3.979221766266502 + 0.000000000000000i
3.918541441830947 + 0.000000000000000i
3.820058967057387 + 0.000000000000000i 
3.684024113506185 + 0.000000000000000i
3.511202932803536 + 0.000000000000000i
3.307439963195127 + 0.000000000000000i
3.080265978640102 + 0.000000000000000i
2.832849552917550 + 0.000000000000000i
2.565972630556613 + 0.000000000000000i
2.283744793210587 + 0.000000000000000i
1.996972474451519 + 0.000000000000000i
0.927737801889518 + 0.670252740725955i
0.927737801889518 - 0.670252740725955i
1.714561796881689 + 0.000000000000000i
-0.015193770830045 + 0.264703483268519i
-0.015193770830045 - 0.264703483268519i
1.438919271663752 + 0.000000000000000i
0.019951101383019 + 0.000000000000000i
0.080534338862828 + 0.000000000000000i 
0.181591307101504 + 0.000000000000000i
0.318955140475174 + 0.000000000000000i
0.488231021129767 + 0.000000000000000i
0.688030188040126 + 0.000000000000000i
1.171318650526539 + 0.000000000000000i
0.917612528393044 + 0.000000000000000i

显然，第二种模式3.642801014184622 + 0.497479798520641i更接近sigma=3.6766133，但ARPACK并没有将其选中。

可能是什么问题？你能帮我搞清楚吗？非常感谢。

Answer 1

关于MATLAB函数的第一件事：

eig返回的特征值为NOT sorted。在[V,D] = eig(A)中，我们只保证V的列是D(i,i)中特征值的对应右特征向量。另一方面，svd返回按降序排序的奇异值。
d = eigs(A,k)返回k 最大幅度的特征值。但是它适用于大型和稀疏矩阵，通常不能替代：
```
d = eig(full(A));
d = sort(d, 'descend');
d = d(1:k);
```
（eigs基于ARPACK，而eig使用LAPACK例程。
There is no natural ordering复数。惯例是sort函数sorts complex elements first by magnitude（即abs(x)），然后是[-pi,pi]区间的相角（即angle(x)），如果幅度相等的话。 / p>

MATLAB

考虑到这一点，请考虑以下MATLAB代码：

% create the same banded matrix you're using
n = 30;
A = spdiags(ones(n,1)*[-1,2,-1], [-1 0 1], n, n);
A(1,9) = 30;
%A = full(A);

% k eigenvalues closest to sigma
k = 10; sigma = 3.6766133;
D = eigs(A, k, sigma);

% lets check they are indeed sorted by distance to sigma
dist = abs(D-sigma);
issorted(dist)

我明白了：

>> D
D =
  3.684024113506185 + 0.000000000000000i
  3.820058967057386 + 0.000000000000000i
  3.511202932803535 + 0.000000000000000i
  3.918541441830945 + 0.000000000000000i
  3.979221766266508 + 0.000000000000000i
  3.307439963195125 + 0.000000000000000i
  4.144524409923134 + 0.000000000000000i
  3.642801014184618 + 0.497479798520640i
  3.642801014184618 - 0.497479798520640i
  3.080265978640096 + 0.000000000000000i

>> dist
dist =
   0.007410813506185
   0.143445667057386
   0.165410367196465
   0.241928141830945
   0.302608466266508
   0.369173336804875
   0.467911109923134
   0.498627536953383
   0.498627536953383
   0.596347321359904

您可以尝试使用密集eig：

获得类似的结果

% closest k eigenvalues to sigma
ev = eig(full(A));
[~,idx] = sort(ev - sigma);
ev = ev(idx(1:k))

% compare against eigs
norm(D - ev)

差异很小（接近机器epsilon）：

>> norm(ev-D)
ans =
     1.257079405021441e-14

的Python

同样在Python中：

import numpy as np
from scipy.sparse import spdiags
from scipy.sparse.linalg import eigs

# create banded matrix
n = 30
A = spdiags((np.ones((n,1))*[-1,2,-1]).T, [-1,0,1], n, n).todense()
A[0,8] = 30

# EIGS: k closest eigenvalues to sigma
k = 10
sigma = 3.6766133
D = eigs(A, k, sigma=sigma, which='LM', return_eigenvectors=False)
D = D[::-1]
for x in D:
    print '{:.16f}'.format(x)

# EIG
ev,_ = np.linalg.eig(A)
idx = np.argsort(np.abs(ev - sigma))
ev = ev[idx[:k]]
for x in ev:
    print '{:.16f}'.format(x)

有类似的结果：

# EIGS
3.6840241135061853+0.0000000000000000j
3.8200589670573866+0.0000000000000000j
3.5112029328035343+0.0000000000000000j
3.9185414418309441+0.0000000000000000j
3.9792217662665070+0.0000000000000000j
3.3074399631951246+0.0000000000000000j
4.1445244099231351+0.0000000000000000j
3.6428010141846170+0.4974797985206380j
3.6428010141846170-0.4974797985206380j
3.0802659786400950+0.0000000000000000j

# EIG
3.6840241135061880+0.0000000000000000j
3.8200589670573906+0.0000000000000000j
3.5112029328035339+0.0000000000000000j
3.9185414418309468+0.0000000000000000j
3.9792217662665008+0.0000000000000000j
3.3074399631951201+0.0000000000000000j
4.1445244099231271+0.0000000000000000j
3.6428010141846201+0.4974797985206384j
3.6428010141846201-0.4974797985206384j
3.0802659786400906+0.0000000000000000j

Answer 2

如果使用eigs函数，NumPy和Matlab之间的结果是一致的：

>> format long
>> A = diag(-ones(n-1,1),-1) + diag(2*ones(n,1)) + diag(-ones(n-1,1),+1);A(1,9)=30;
>> eigs(A,3,3.6766133)'
ans =
   3.684024113506185   3.820058967057386   3.511202932803534

至于为什么没有选择真正最接近的特征值，我认为这与收敛到实矩阵的复特征值和实际移位的选择有关。我不知道ARPACK如何计算其迭代，但我记得被告知具有实σ的实数A不能默认收敛到复共轭对，因为它们的绝对值比为1（对于逆功率迭代）。由于ARPACK将在8 ^th和9 ^th迭代中生成复杂的特征值（+/-排序是随机的），我猜他们是对此的一些修复我已经忘记或永远不会知道：

>> ev = eigs(A,9,3.6766133);ev(8:9)
ans =
  3.642801014184617 - 0.497479798520639i
  3.642801014184617 + 0.497479798520639i

我不确定它们是否是一个普遍的工作，除了猜测移位的复杂部分或只是抓住额外的特征值，直到共轭对落入ARPACK方法的收敛球。

使用ARPACK解决特征值问题，但使用Matlab获得不一致的结果

2 个答案:

MATLAB

的Python