Why can't I solve eigenvector correctly? (eigenvalue problem)

The following matrix H is composed of two different constant a and b:

 a, b, 0, 0, 0, b
 b, a, b, 0, 0, 0
 0, b, a, b, 0, 0
 0, 0, b, a, b, 0
 0, 0, 0, b, a, b
 b, 0, 0, 0, b, a

When I enter [X,E] = eig(H), it returns me:

X =

 -1,  1, -1, -1, -1, 1
 -1,  0,  1,  0,  1, 1
  0, -1,  0,  1, -1, 1
  1, -1, -1, -1,  1, 1
  1,  0,  1,  0, -1, 1
  0,  1,  0,  1,  1, 1

E =

 a + b,     0,     0,     0,       0,       0
     0, a + b,     0,     0,       0,       0
     0,     0, a - b,     0,       0,       0
     0,     0,     0, a - b,       0,       0
     0,     0,     0,     0, a - 2*b,       0
     0,     0,     0,     0,       0, a + 2*b

I first thought this was right, but then I put actual number a = -11.7 and b = -0.7 and then I get

X =

   -0.4082   -0.5000    0.2887   -0.5000    0.2887    0.4082
   -0.4082   -0.5000   -0.2887    0.5000    0.2887   -0.4082
   -0.4082    0.0000   -0.5774   -0.0000   -0.5774    0.4082
   -0.4082    0.5000   -0.2887   -0.5000    0.2887   -0.4082
   -0.4082    0.5000    0.2887    0.5000    0.2887    0.4082
   -0.4082         0    0.5774         0   -0.5774   -0.4082

E =

-13.1000         0         0         0         0         0
       0  -12.4000         0         0         0         0
       0         0  -12.4000         0         0         0
       0         0         0  -11.0000         0         0
       0         0         0         0  -11.0000         0
       0         0         0         0         0  -10.3000

Now if I enter a = -11.7 and b = -0.7 AFTER solving the eigenvalue problem, the Eigenvalue are still the same (except the order is different). However, the eigenvectors seems to be not consistent with each other.

Why am I not getting the right general solution when solving symbolically, and what can I do to do it correctly?

Answer 1

[Steven Lord]

Any linear combination of eigenvectors (ignoring the all-zero vector) associated with an eigenvalue q is also an eigenvector associated with the eigenvalue q. So while this vector:

v1 = repmat(-0.4082, 6, 1);

looks very different from this vector:

v2 = -ones(6, 1);

they're both eigenvectors for the eigenvalue a + 2*b (which is -13.1000 in your numeric example.)

To take a look at something a little more challenging, while the fourth and fifth columns of the X matrix from your numerical calculations doesn't look like the third and fourth columns of the X matrix from your symbolic calculations they are all eigenvectors associated with the eigenvalue a-b (or -11 in your numeric example.)

Specifically, column 4 of the numeric X is 0.5 times column 3 of the symbolic X. Column 5 of the numeric X is 0.2887 times column 3 of the symbolic X minus 0.5774 times column 4 of the symbolic X.

So all the answers you received are correct, they just look a little different than you expected.

[Hayao]

Thanks for the answer. Now that you say it, I can see it too.

However, this would be quite problematic for me. This eigenvalue problem actually comes from Huckel Molecular Orbitals Method for benzene. For benzene, the eigenvector for symbolic result must look like the result achieved for numeric solution, that is, all of the elements vn in an eigenvector must be normalized to meet the condition:

   |v1|^2 + |v2|^2 + |v3|^2 + |v4|^2 + |v5|^2 + |v6|^2 = 1

If not, then that would be contradicting result because the matrix element of the matrix H has elements constructed normalized (they were constructed from complete orthonormal basis).

Also, especially for column 3 and 5 as eigenvector for numeric solution, they must be that particular element, or else it would mean that molecular orbital would be the same with another orbital (which is impossible), in this case eigenvector for column 2 and 4.

Finally, they should also be in the particular order they are in for the symbolic solution.

Is there a workaround to do get symbolic solution to look exactly like numeric solution (for eigenvectors)?