Are eigenvectors returned by R function eigen() wrong?

Question

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

eigen(a)

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?

Answer 1

Some paper work tells you

• 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.

Scaling eigenvectors to unit-length gives

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

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

---

How to find eigenvectors using textbook method

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

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