MATLAB中的广义特征向量?

问题描述

在具有单个或至少很少几个命令的高特征值多重性的情况下，是否存在一种获取广义特征向量的方法?如果每个特征值的多重性为1，我可以使用[V,D] = eig(A)，但是此命令不适用于多个特征值.

Is there a way to obtain generalized eigenvectors in case of high multiplicity of eigenvalues with a single one or at least very few commands ? In case of multiplicity 1 for each eigenvalue I can use [V,D] = eig(A), but this command doesn't work for multiple eigenvalues.

推荐答案

根据Matlab文档，[V，D] = eig(A，B)生成具有广义特征值的对角矩阵D和列为相应的特征向量，使得A * V = B * V * D

According to Matlab documentation, [V,D] = eig(A,B) produces a diagonal matrix D of generalized eigenvalues and a full matrix V whose columns are the corresponding eigenvectors so that A*V = B*V*D

下面是一个如何自己做的示例...首先，我们输入一个样本矩阵A:

Here an example how to do it yourself... First we enter a sample matrix A:

 A = [ 35  -12   4   30 ;
       22   -8   3   19 ;
      -10    3   0   -9 ;
      -27    9  -3  -23 ];

然后，我们研究其特征多项式，特征值和特征向量.

Then we explore its characteristic polynomial, eigenvalues, and eigenvectors.

 poly(A)
 ans =
     1.0000   -4.0000    6.0000   -4.0000    1.0000

这些是特征多项式的系数，因此为(λ− 1)^ 4然后

These are the coefficients of the characteristic polynomial, which hence is (λ − 1)^4Then

 [V, D] = eigensys(A)
 V =
 [ 1, 0]
 [ 0, 1]
 [-1, 3]
 [-1, 0]


 D =
 [1]
 [1]
 [1]
 [1]

因此MATLAB仅找到两个独立的特征向量

Thus MATLAB finds only the two independent eigenvectors

 w1 = [1  0  -1  -1]';
 w2 = [0  1   3   0]';

与单个多重性4特征值λ= 1相关，因此具有缺陷2.
因此，我们建立了4x4单位矩阵，矩阵B =A-λI

associated with the single multiplicity 4 eigenvalue λ=1 , which therefore has defect 2.
So we set up the 4x4 identity matrix and the matrix B=A-λI

  Id = eye(4);
  B = A - L*Id;

当L = 1时，我们计算B ^ 2和B ^ 3

with L=1, When we calculate B^2 and B^3

  B2 = B*B
  B3 = B2*B

我们发现B2≠0，但B3 = 0，因此应该有一条与之相关的长度为3的链
特征值λ= 1.选择第一个广义特征向量

We find that B2 ≠ 0, but B3 = 0, so there should be a length 3 chain associated with
the eigenvalue λ = 1 . Choosing the first generalized eigenvector

 u1 = [1  0  0  0]';

我们计算出进一步的广义特征向量

we calculate the further generalized eigenvectors

 u2 = B*u1
 u2 =
     34
     22
    -10
    -27

和

 u3 = B*u2
 u3 =
     42
      7
    -21
    -42

因此，我们根据(普通)找到了长度为3的链{u3，u2，u1}特征向量u3. (要使该结果与MATLAB的eigensys计算相一致，您可以可以检查u3-42w1 = 7w2)

Thus we have found the length 3 chain {u3, u2, u1} based on the (ordinary)eigenvector u3. (To reconcile this result with MATLAB's eigensys calculation, youcan check that u3-42w1=7w2)

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