(adsbygoogle = window.adsbygoogle || []).push({}); 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 = A^{T}

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-[itex]\lambda[/itex]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 [itex]\lambda[/itex]'s in the diagonal.

Thus, is there anyway we can look at the canonical form of A-[itex]\lambda[/itex]I and deduce from it's rank that it has no multiple eigenvalues?

B) The hint says to use a similar strategy as A

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# Homework Help: Symmetric, irreducible, tridiagonal matrix: Eigenvalues

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