If you can use exact rational numbers as input, Mathematica can do the calculation exactly (as you saw originally). So, use 11/10 instead of 1.1. If you must use decimal points, get the rational representation using Rationalize[].

It appears to be entirely numeric, whether exact or approximate. Given the difference in scales of the inputs, I would not expect a computation at machine precision to do well in such circumstances. The scale factor of 10 and the error of around 10^(-6) are consistent with working at 16 digits of precision (machine double, that is).

Hi Jonathan, in fact the matrix is rather simple, so you can solve it as a symbolic matrix:

(mat = {{0, 0, 0, a, -I*a, b}, {0, 0, 0, a, -I*a, b}, {0, 0, 0, 
     a, -I*a, b}, {a, a, a, c, 0, 0}, {-I*a, -I*a, -I*a, 0, c, 0}, {b,
      b, b, 0, 0, c}}) // MatrixForm

{ev, evec} = Eigensystem[mat];
ev
{0, 0, c, c, 1/2 (c - Sqrt[12 b^2 + c^2]), 
 1/2 (c + Sqrt[12 b^2 + c^2])}

Replacing the symbols with exact numbers results in:

s1 = ev /. { a -> 1, b -> 100, c -> 10^10}
{0, 0, 10000000000, 10000000000, 
 1/2 (10000000000 - 200 Sqrt[2500000000000003]), 
 1/2 (10000000000 + 200 Sqrt[2500000000000003])}

Replacing the symbols with numeric arguments gives:

s2 = ev /. { a -> 1., b -> 100., c -> 1. 10^10}
{0, 0, 1.*10^10, 1.*10^10, -2.86102*10^-6, 1.*10^10}