# Eigenvalues of a Householder matrix

• Oct 7th 2011, 02:14 AM
davesface
Eigenvalues of a Householder matrix
We will prove in class that for all of the eigenvalues of an n × n orthogonal matrix, $|\lambda_i|^2$ = 1, and in general will be complex. Consider here the eigenvalue problem, Hx = $\lambda$x, for a Householder matrix.

(a) Starting with the fact that $|\lambda_i|$ $^2$ = 1, how do we know that all eigenvalues of H are either +1 or (−1)?

(b) Beginning with Hx = $\lambda$x, show that there are only two possibilities, $\lambda_i$ = −1, or $v^T e_i = 0$, for all i = 1, 2, . . . n (*), where e_i is an eigenvector.

(c) In general, only one eigenvalue can be −1. That is, for and n × n Householder matrix, one eigenvalue is (−1), and there is (n − 1) algebraic and geometric multiplicity for $\lambda$ = 1. Using (*) above, explain why this is so.
• Sep 20th 2014, 07:11 PM
phys251
Re: Eigenvalues of a Householder matrix
(a) All eigenvalues of A are $\pm 1$ whenever A is unitary and Hermetian. Prove the Hermetian part of this (they already took care of the unitary part).

(b) not sure...

(c) I saw the answer but was not able to figure out a key part of the proof. If we let $\mathbf{x} = \mathbf{v},$ we get $F\mathbf{v} = -\mathbf{v}.$ That means that exactly one of the eigenvalues is -1, but I am not sure why.