R 函数 eigen() 返回的特征向量是错误的吗? [英] Are eigenvectors returned by R function eigen() wrong?

问题描述

#eigen values and vectors
a <- matrix(c(2, -1, -1, 2), 2)

eigen(a)

我试图在 R 中找到特征值和特征向量.函数 eigen 适用于特征值，但特​​征向量值存在错误.有什么办法可以解决吗?

I am trying to find eigenvalues and eigenvectors in R. Function eigen works for eigenvalues but there are errors in eigenvectors values. Is there any way to fix that?

推荐答案

一些文书工作告诉你

• 特征值 3 的特征向量是 (-s, s) 对于任何非零实数值 s;
• 对于任何非零实数值 t，特征值 1 的特征向量是 (t, t).

• the eigenvector for eigenvalue 3 is (-s, s) for any non-zero real value s;
• the eigenvector for eigenvalue 1 is (t, t) for any non-zero real value t.

将特征向量缩放到单位长度给出

Scaling eigenvectors to unit-length gives

s = ± sqrt(0.5) = ±0.7071068
t = ± sqrt(0.5) = ±0.7071068

缩放是好的，因为如果矩阵是实对称的，特征向量的矩阵是正交的，所以它的逆是它的转置.以您的真实对称矩阵 a 为例:

Scaling is good because if the matrix is real symmetric, the matrix of eigenvectors is orthonormal, so that its inverse is its transpose. Taking your real symmetric matrix a for example:

a <- matrix(c(2, -1, -1, 2), 2)
#     [,1] [,2]
#[1,]    2   -1
#[2,]   -1    2

E <- eigen(a)

d <- E[[1]]
#[1] 3 1

u <- E[[2]]
#           [,1]       [,2]
#[1,] -0.7071068 -0.7071068
#[2,]  0.7071068 -0.7071068

u %*% diag(d) %*% solve(u)  ## don't do this stupid computation in practice
#     [,1] [,2]
#[1,]    2   -1
#[2,]   -1    2

u %*% diag(d) %*% t(u)      ## don't do this stupid computation in practice
#     [,1] [,2]
#[1,]    2   -1
#[2,]   -1    2

crossprod(u)
#     [,1] [,2]
#[1,]    1    0
#[2,]    0    1

tcrossprod(u)
#     [,1] [,2]
#[1,]    1    0
#[2,]    0    1

<小时>

如何用教科书的方法求特征向量

教科书的方法是求解齐次系统:(A - λI)x = 0 为 Null Space 基.我的这个答案中的 NullSpace 函数会很有帮助.

The textbook method is to solve the homogenous system: (A - λI)x = 0 for the Null Space basis. The NullSpace function in my this answer would be helpful.

## your matrix
a <- matrix(c(2, -1, -1, 2), 2)

## knowing that eigenvalues are 3 and 1

## eigenvector for eigenvalue 3
NullSpace(a - diag(3, nrow(a)))
#     [,1]
#[1,]   -1
#[2,]    1

## eigenvector for eigenvalue 1
NullSpace(a - diag(1, nrow(a)))
#     [,1]
#[1,]    1
#[2,]    1

如您所见，它们并未标准化".相比之下，pracma::nullspace 给出了标准化"的特征向量，所以你得到的东西与 eigen 的输出一致(直到可能的符号翻转):

As you can see, they are not "normalized". By contrasts, pracma::nullspace gives "normalized" eigenvectors, so you get something consistent with the output of eigen (up to possible sign flipping):

library(pracma)

nullspace(a - diag(3, nrow(a)))
#           [,1]
#[1,] -0.7071068
#[2,]  0.7071068

nullspace(a - diag(1, nrow(a)))
#          [,1]
#[1,] 0.7071068
#[2,] 0.7071068

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