# Determinant of this special matrix

1. Jun 2, 2010

### ian2012

Hope someone can help me out here.. I've found the eigenvalues (lamda) of this matrix, but through a very very long way, does anyone know of a quicker way (there must be a quicker way). The matrix is 3x3:

H = ( E, a, a; a, E, a; a, a, E ).

I can reduce the determinant to the following, but still troublesome:

det(H) = | E - lamda, a, a; a, E - lamda, a; a, a, E - lamda |

det(H) = | E - lamda - a, a - E + lamda, 0; 0, E - lamda - a, a - E + lamda; a, a, E - lamda |

2. Jun 2, 2010

### Filip Larsen

For what its worth, Mathematica gives E-a and 2a+E as a double and single eigenvalue of H, respectively. But using Mathematica is probably not what you meant by "a quicker way".

Does following through on the calculation you indicated in your post give you any problems?

3. Jun 2, 2010

### ian2012

Yeah, i know the solutions , thanks... Well i end up with (E - lamda - a)(E - lamda) + a(E - lamda - a) - a = 0 as the equation of the determinant. I am a bit rusty on my high school algebra, is there a way to factorise that? (You could multiply it all out and spend a lifetime dividing it by a trial solution)

Last edited: Jun 2, 2010
4. Jun 2, 2010

### ian2012

Okay, i may have cancelled abit too early, the factorization I am looking for is of:
(E - lamda - a)^2(E- lamda + a) = a(E - lamda - a)

5. Jun 2, 2010

### Dickfore

The eigenvalue problem is equivalent to:
$$E \, x_{n} + a \, \sum_{m \ne n}{x_{m}} = \lambda \, x_{n}$$

If you introduce:
$$S = \sum_{m}{x_{m}}$$

you can rewrite the equations as:
$$(E - a - \lambda) \, x_{n} + a \, S = 0$$

$$x_{n} = \frac{a \, S}{\lambda + a - E}$$
If $S = 0$, then, in order that we have at least one non-zero $x_{n}$, we must have:

$$\lambda + a - E = 0$$

or

$$\lambda = E - a$$

If, on the other hand $S \ne 0$, we have unique solutions for $x_{n}$. Substituting these back into the definition for $S$ and canceling, we get:

$$S = \sum_{n}{\frac{a \, S}{\lambda + a - E}} = \frac{N \, a\, S}{\lambda + a - E}$$

$$1 = \frac{a \, N}{\lambda + a - E}$$

$$\lambda = E + (N - 1) \, a$$

BTW, this problem is similar to the Cooper pairing problem in superconductivity.

6. Jun 2, 2010

### kabindra

Following are the eigenvectors by inspection: [1. 1,1]', [1 -1 0]' and [0 1 -1]'. These are linearly independent with the corresponding eigenvalues E+2a, E-a and E-a.

7. Jun 3, 2010

### Simon_Tyler

@Dickfore:
That's a nice way of approaching the problem.