How do I determine orthogonal eigenvectors for a repeated eigenvalue?

[J.Asher]

Homework Statement

There is an Hamiltonian operation which is given by
(2 1 1)
(1 2 1) = H ; 3-by-3 matrix
(1 1 2)
And let's have an arbtrary eigenvector
(a)
(b) = v ; (3x1) matrix
(c)

Then, from the characteristic equation, the eigenvalues are 1,4. Here eigenvalue 1 is
repeated one.

Homework Equations

Now, my question arises. I know that the eigenvectors that corresponds to eigenvalue 1 is two and both are orthgonal to each other. However, I can't find any of them because when
I substitute eigenvalue into 1, I get a kind of meaningless(?) equation,

(2 1 1) (a) (a)
(1 2 1) (b) = (b)
(1 1 2) (c) (c) .

And it gives 3-equivalent equation, a+b+c = 0
I couldn't determine any relation between a,b,c.

The Attempt at a Solution

So I quess that do I have to choose a,b,c satisfying the condition a+b+c=0
but still problem doesn't disappear beacuse there would be a lot of soulutions...

What did I wrong?

 

[Mark44]

You didn't take it far enough.

a + b + c = 0 ==>
a = -b - c
b = b
c = ... c

The eigenvectors form a two-dimensional subspace of R³ that is spanned by two vectors. If you look closely at the system above, you can pick out the two vectors. These two vectors are linearly independent but they don't happen to be orthogonal.

 

[J.Asher]

The operator we handle is Hermitian since Hamlitonian is real physical operator.
it is no doubt that the eigenvectors of the Hermitian is orthogonal.
Actually I cannot understand what you are saying. What do you mean by that?
...??

 

[Dick]

The operator we handle is Hermitian since Hamlitonian is real physical operator.
it is no doubt that the eigenvectors of the Hermitian is orthogonal.
Actually I cannot understand what you are saying. What do you mean by that?
...??

If it's hermitian the eigenvectors belonging to distinct eigenvalues are orthogonal. Eigenvectors belonging to the same eigenvalue don't have to be. You have to select them to be orthogonal.