问题描述
我的问题的简短版本:
如果我们已经知道属于特征向量的特征值，那么为矩阵A计算特征向量的最佳方法是什么?
What would be the optimal way of calculating an eigenvector for a matrix A, if we already know the eigenvalue belonging to the eigenvector?
详细说明:
我有一个大的随机矩阵A，因为它是随机的，所以具有非负的左特征向量x(即A^Tx=x).
I have a large stochastic matrix A which, because it is stochastic, has a non-negative left eigenvector x (such that A^Tx=x).
我正在寻找一种快速有效的数值计算此向量的方法. (最好在MATLAB或numpy/scipy中使用-由于这两种方法都围绕ARPACK/LAPACK，所以任何一种都可以).
I'm looking for quick and efficient methods of numerically calculating this vector. (Preferrably in MATLAB or numpy/scipy - since both of these wrap around ARPACK/LAPACK, any one would be fine).
我知道1是A的最大特征值，所以我知道调用类似以下Python代码的代码:
I know that 1 is the largest eigenvalue of A, so I know that calling something like this Python code:
from scipy.sparse.linalg import eigs
vals, vecs = eigs(A, k=1)
将导致vals = 1和vecs等于我需要的向量.
will result in vals = 1 and vecs equalling the vector I need.
但是，令我困扰的是，通常，计算本征值比求解线性系统要困难得多，而且，通常，如果矩阵M具有本征值l，则找到适当的特征向量是求解方程式(M - 1 * I) * x = 0的问题，因为从理论上讲，这仅是求解线性系统，更具体地说，是找到矩阵的零空间，所以从理论上讲，这比计算特征值更简单.
However, the thing that bothers me here is that calculating eigenvalues is, in general, a more difficult operation than solving a linear system, and, in general, if a matrix M has eigenvalue l, then finding the appropriate eigenvector is a matter of solving the equation (M - 1 * I) * x = 0, which is, in theory at least, an operation that is simpler than calculating an eigenvalue, since we are only solving a linear system, more specifically, finding the nullspace of a matrix.
但是，我发现MATLAB中的所有空空间计算方法都依赖于svd计算，这是我无法承受的对我大小的矩阵执行的过程.我也不能在线性方程上调用解算器，因为它们都只能找到一个解，而该解是0(是的，这是一个解，但不是我需要的一个解).
However, I find that all methods of nullspace calculation in MATLAB rely on svd calculation, a process I cannot afford to perform on a matrix of my size. I also cannot call solvers on the linear equation, because they all only find one solution, and that solution is 0 (which, yes, is a solution, but not the one I need).
与通过计算最大特征值和随附的特征向量相比，有什么方法可以避免调用类似eigs的函数来更快地解决我的问题?
Is there any way to avoid calls to eigs-like function to solve my problem more quickly than by calculating the largest eigenvalue and accompanying eigenvector?
推荐答案
以下是使用Matlab的一种方法:
Here's one approach using Matlab:
- 让 x 表示与特征值1相关的(行)左特征向量.它满足线性方程(或矩阵方程)系统 xA = x ，或 x ( A - I )= 0 .
- 为避免对该方程组进行全零解，请删除第一个方程，并在其余方程中将 x 的第一个条目任意设置为1.
- 求解剩余的方程式( x = 1)以获得 x 的其他项.
- Let x denote the (row) left eigenvector associated to eigenvalue 1. It satisfies the system of linear equations (or matrix equation) xA = x, or x(A−I)=0.
- To avoid the all-zeros solution to that system of equations, remove the first equation and arbitrarily set the first entry of x to 1 in the remaining equations.
- Solve those remaining equations (with x = 1) to obtain the other entries of x.
使用Matlab的示例:
Example using Matlab:
>> A = [.6 .1 .3
.2 .7 .1
.5 .1 .4]; %// example stochastic matrix
>> x = [1, -A(1, 2:end)/(A(2:end, 2:end)-eye(size(A,1)-1))]
x =
1.000000000000000 0.529411764705882 0.588235294117647
>> x*A %// check
ans =
1.000000000000000 0.529411764705882 0.588235294117647
请注意，代码-A(1, 2:end)/(A(2:end, 2:end)-eye(size(A,1)-1))是步骤3.
Note that the code -A(1, 2:end)/(A(2:end, 2:end)-eye(size(A,1)-1)) is step 3.
在您的公式中，您定义 x 为 A (例如 A x = x ).这只是上面代码中的x.':
In your formulation you define x to be a (column) right eigenvector of A (such that Ax = x). This is just x.' from the above code:
>> x = x.'
x =
1.000000000000000
0.529411764705882
0.588235294117647
>> A.'*x %// check
ans =
1.000000000000000
0.529411764705882
0.588235294117647
您当然可以对特征向量进行归一化来求和1:
You can of course normalize the eigenvector to sum 1:
>> x = x/sum(x)
x =
0.472222222222222
0.250000000000000
0.277777777777778
>> A.'*x %'// check
ans =
0.472222222222222
0.250000000000000
0.277777777777778
遵循通常的惯例.等效地，这对应于转置的矩阵的 right 特征向量.
Following the usual convention. Equivalently, this corresponds to a right eigenvector of the transposed matrix.
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