R中不同特征向量的数量 [英] number of distinct eigenvectors in R

问题描述

在下面的示例中，有2个特征向量对应于1个特征值(多重性为4).但是，R返回4个不同的特征向量.看起来它们之间的对几乎相同，只是机器浮点误差(ε)不同.您可以检查并确认吗?

There are 2 eigenvectors corresponding to 1 eigenvalue (with multiplicity 4) for the following example. However, R returns 4 distinct eigenvectors. It looks like pairs of them are approximately the same only differing in machine floating point error (epsilon). Can you please check and confirm?

> B
 [,1] [,2] [,3] [,4]
 [1,]    2    0    0    0
 [2,]    1    2    0    0
 [3,]    0    1    2    0
 [4,]    0    0    1    2
> eigen(B)
$values
[1] 2 2 2 2

$vectors
[,1]          [,2]          [,3]          [,4]
[1,]    0  0.000000e+00  0.000000e+00  8.758115e-47
[2,]    0  0.000000e+00  1.972152e-31 -1.972152e-31
[3,]    0  4.440892e-16 -4.440892e-16  4.440892e-16
[4,]    1 -1.000000e+00  1.000000e+00 -1.000000e+00

推荐答案

以下是答案:

http://www.wolframalpha.com/input/?i=eigenvalues+[[2,+0,+0,+0],+[1,+2,+0,+0],+[0,+1,+2,+0],+[0,+0,+1,+2]]

(我无法建立链接...)

(I cannot make it to a link...)

这样想:

      2    0    0    0
B  =  1    2    0    0
      0    1    2    0
      0    0    1    2.

如果我们从主对角线中减去特征值\ lambda = 2(就像计算特征空间一样)，则得出

If we subtract the eigenvalue \lambda = 2 from the main diagonal (as one does computing eigenspaces), we obtain

              0    0    0    0
(B - 2 I)  =  1    0    0    0
              0    1    0    0
              0    0    1    0.

如果坐标为(x，y，z，w)，那么显然(B-2 I)X = 0得出x = 0(第二行)，y = 0(第三行)，并且z = 0(从最后一行开始).因此，空间由所有点(0，0，0，w)组成，其中w是任意的.也就是说，它是一维的，任何矢量(0、0、0，t)都将用作基本矢量(t非零).

If the coordinates are (x, y, z, w), then, obviously (B - 2 I) X = 0 yields x = 0 (from the second row), y = 0 (from the third row), and z = 0 (from the last row). Hence the space consists of all points (0, 0, 0, w) where w is arbitrary. That is, it is one-dimensional and any vector (0, 0, 0, t) will serve as a basis vector (t nonzero).

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