# Symmetric, irreducible, tridiagonal matrix: Eigenvalues

1. Feb 11, 2012

### Scootertaj

1. The problem statement, all variables and given/known data
A) Let A be a symmetric, irreducible, tridiagonal matrix. Show that A cannot have a multiple eigenvalue.
B) Let A be an upper Hessenberg matrix with all its subdiagonal elements non-zero. Assume A has a multiple eigenvalue. Show that there can only be one eigenvector associated with it.

2. Relevant definitions
Symmetric: A = AT
Irreducible: not a good definition for it, but essentially it's well-connected
Tridiagonal: obvious
Upper Hessenberg: http://en.wikipedia.org/wiki/Hessenberg_matrix Essentially, it's a mix between a tridiagonal and a upper triangular. It's tridiagonal with non-zero elements above its 3rd subdiagonal.

3. The attempt at a solution
A) It seems contradiction would be the best combined with using the Jordan canonical form of A-$\lambda$I.
Jordan canonical form: http://en.wikipedia.org/wiki/Jordan_normal_form
So, we know that we can get the Jordan canonical form of A which has all $\lambda$'s in the diagonal.
Thus, is there anyway we can look at the canonical form of A-$\lambda$I and deduce from it's rank that it has no multiple eigenvalues?
B) The hint says to use a similar strategy as A

2. Feb 12, 2012

### Scootertaj

Just a bump.

3. Feb 13, 2012

### Scootertaj

Just another bump. This is a tough problem.