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

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RMT on 12 Jan 2020
Edited: David Goodmanson on 12 Jan 2020
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

David Goodmanson on 12 Jan 2020
Edited: David Goodmanson on 12 Jan 2020
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.