Why did i get different answers for eigenvectors for different powers of a matrix?

A= [1 0 1; 0 0 0; 1 0 -1]

[U,D] = eig(A^1);

The output:

D: eigenvalues

-1.4142 0 0

0 0 0

0 0 1.4142

U: eigenvectors

0.3827 0 0.9239

0 -1.0000 0

-0.9239 0 0.3827

I was doing the calculations by hand so i thought it's easier to calculate the eigenvectors for A^2 because eigenvalues will be {0,2,2} and i should get the same eigenvectors of A since they doen't change for matrix powers. And i wanted to confirm this by MATLAB but i got different answers. Any explanation?

[U,D] = eig(A^2);

D: eigenvalues (correct)

0 0 0

0 2 0

0 0 2

U: eigenvectors( here is my problem why they have changed?)

0 1 0

1 0 0

0 0 1

Answer 1

[David Goodmanson]

Hi RMT,

Since you have two identical eigenvalues, the eigenvectors corresponding to those eigenvalues are not unique*. They only need to span the subspace that is defined by the two of them, so in that subspace the columns of

[1 0
 0 1]

. qualify as eigenvectors, as do the columns of, for example

[1/sqrt(2) 1/sqrt(2)
[-1/sqrt(2) 1/sqrt(2)]

eig has no requirement on the order of the eigenvalues it generates. It's not helping that eig(A) has the zero eigenvalue second, and eig(A^2) has the zero eigenvalue first That obscures where the subspaces are. However, in the A case you can eliminate the second column since it corresponds to lambda = 0, and in the A^2 case you can eliminate the first column for the same reason. That leaves

 0.3827   0.9239
    0       0
-0.9239     0.3827

and 
1     0
0     0
0     1

Neither of these involves the second coordinate, so if you toss that out you are comparing

 0.3827   0.9239
-0.9239   0.3827

with 
1     0
0     1

so you can see it's just a rotation in the 2d space corresponding to the two degenerate eigenvalues.

*this does not count the fact that technically if v is an eigenvector then so is const*v, i.e. eigenvector normalization is a separate requirement. But normalization only changes the length, not the direction.