Determinant of Gram matrix is non-zero, but vectors are not linearly independent

Question

From Wikipedia:

a set of vectors is linearly independent if and only if the Gram determinant (the determinant of the Gram matrix) is non-zero.

But consider the matrix M:

  5   4   2   1
  0   1  -1  -1
 -1  -1   3   0
  1   1  -1   2

The Gram matrix G is:

 27  22   6  7
 22  19   3  5
  6   3  15  1
  7   5   1  6

And the determinant of G is 1024. The eigenvalues of M are:

4, 4, 1, 2

The eigenvectors of M are:

 -0.58   0.58  -0.71  -0.58
  0.0   -0.0    0.71   0.58
  0.58  -0.58   0.0    0.0
 -0.58   0.58   0.0   -0.58

And the rank of this eigenvector matrix is 3, not 4, so... the vectors are not linearly independent. Am I missing something obvious here? Thanks.

Answer 1

Consider the matrix

1  1
0  1

It's got one eigenvector,

0
1

so its eigenvector matrix, as reported by matlab, will be

0  0
1  1

Its eigenvalues are 1 and 1. But it's got rank 2, even though the eignevector matrix has rank 1. So what you're missing is a clear understanding of what matlab returns when it computes an eigenvector matrix.