MATLAB: Can’t I solve eigenvector correctly? (eigenvalue problem)

matrix eigenvalue eigenvector problem symbolic

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?

Best Answer

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.
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